The solution for Postgres is to use substring() substring('one two three four five' from '(\S+)') yields one. The biggest difference to Oracle's regexp_substring() is that you can't locate the "nth" occurance directly. To get the second word, you could use: substring('one two three four five' from '\S+\s*(\S+)') If you are trying to access words (separated by whitespace), then splitting the string into an array is another alternative: (regexp_split_to_array('one two three four five', '\s+'))[1] The above returns the first word, [2] would return the second. The regex defines the separator (delimiter) between each element. Note that Postgres 15 will have a regexp_substr() that is pretty much identical to Oracle's implementation.

It's possible to use https://jmeter.apache.org/usermanual/functions.html in https://jmeter.apache.org/usermanual/component_reference.html#JSON_Extractor https://github.com/json-path/JsonPath#path-examples expressions, your syntax is simply incorrect (you can check it yourself using https://www.blazemeter.com/blog/how-debug-your-apache-jmeter-script ) My expectation is that you should be something like: $..valid_model_list[?(@.model_name =~ /.*Model-Corpus-${variable}-${counter}-v1-8k/i)] Demo:

Just place a space inside the brackets:/^(?=.*?[a-z])(?=(?:.*?\d){9})[a-z\d ]{12,}$/i ^ aqui Notice you have a space between \d and ].The brackets define a https://www.regular-expressions.info/charclass.html , that is, take any character inside them.In the case, [a-z\d ], means "a letter of a a z, or a \d (digit from 0 to 9), or a space".

Notice that the spaces are right within your string:yaml_content = """ MAIN : PROD : {} LOCATION : {} EXTRA : LOLO : {} """.format(PROD, LOCATION, LOLO) And the easiest way to solve this would be to produce something like:yaml_content = ( "MAIN:\n"+ "\tPROD: {}\n" + "\tLOCATION: {}\n" + "EXTRA:\n" + "\tLOLO: {}\n").format(PROD, LOCATION, LOLO) No need to use regex to such.Of course, if what you want is to create a YAML file from data already stored in variables I recommend that you use the module https://pyyaml.org/ because it will just put the data inside a dictionary and the language will do the rest for you.

To look for patterns in a text, a good solution is to use regular expressions. In Python, they are available in https://docs.python.org/3/library/re.html .To search for the numbers, you can use the https://www.regular-expressions.info/shorthand.html \d (which takes any digit from 0 to 91). And to limit the amount, you can use the https://www.regular-expressions.info/repeat.html#limit , how {2} and {4}, which mean, respectively, "2 occurrences" and "4 occurrences". That is, \d{2} means "2 occurrences of digits from 0 to 9".So the regex is like this:import re texto = """Data 01.02.2019, outra data 20.11.2018 etc... Outra data 15.03.1980, etc """ r = re.compile(r'\d{2}.\d{2}.\d{4}') print(r.findall(texto)) https://docs.python.org/3/library/re.html#re.findall returns a list with the sections found in the string. In the case, it is any excerpt that corresponds to \d{2}.\d{2}.\d{4} (two digits, point, two digits, point, four digits):['01.02.2019', '20.11.2018', '15.03.1980']Notice that the point was written as .. This is necessary because the point has special meaning in regex: https://docs.python.org/3/library/re.html#index-0 (except https://pt.stackoverflow.com/q/372844/112052 ). That is, if the regex was \d{2}.\d{2}.\d{4}, she would take things like 12-10#2018 or even 12a1092018, https://ideone.com/3dTBSM (the point corresponds to any character, including letters and numbers).In order for the point to "lost its powers" and to be interpreted as a common character, one must do the escape escape with </code>. So, . corresponds to only the character .without any special significance.If you want, you can also use https://docs.python.org/3/library/re.html#re.finditer , which returns one iterator of https://docs.python.org/3/library/re.html#match-objects , which can be used to get more information about the sections found:for match in r.finditer(texto): print("data '{}' encontrada na posicao {}".format(match.group(), match.start())) In the above example I used https://docs.python.org/3/library/re.html#re.Match.group to get the very excerpt that was found (in the case, the date) and https://docs.python.org/3/library/re.html#re.Match.start , which returns the position of the string in which the date was found:date '01.02.2019' found in position 5date '20.11.2018' found in heading 28date '15.03.1980' found in heading 57See https://docs.python.org/3/library/re.html#match-objects for more details about the information that can be obtained from match.We could stop here, but since it's not clear what you can have in your text, I think we can improve this regex a little more.Limiting accepted values\d search for any digit from 0 to 9, which means that \d{2} will accept values as 00, 32 and 99. But these are not valid values for days and months, so we can change the regex to limit the accepted values:r = re.compile(r'(0[1-9]|[12]\d|3[01]).(0[1-9]|1[0-2]).(19|20)\d{2}') Here we have the use of https://www.regular-expressions.info/alternation.html (the character |, which means "or"). That is, abc|xyz means "abc or xyz". In the above regex, we have several cases so to cover several possibilities of values. For example, for the day, we have 3 possibilities:0[1-9]: a zero, followed by "a digit of 1 to 9". The brackets define a https://www.regular-expressions.info/charclass.html , and the hyphen defines a range. So, [1-9] defines a character that can be any digit from 1 to 9. This ensures that the day can be 01 a 09[12]\d: [12] is also a character class, but without the hyphen (and therefore without a range). In the case, it means "the digit 1 or 2". Then all this passage means "digit 1 or 2, followed by any digit". This ensures that the day can be 10 a 293[01]: The digit 3 followed by 0 or 1 (for the days 30 and 31)O | between these 3 expressions ensures that regex can take any of these possibilities, and the parentheses around all this group all this sub-expression into one thing only.Something similar was done with the month, and 0[1-9]|1[0-2] means:0[1-9]: zero, followed by a digit from 1 to 9 (for the months of 01 a 09), or1[0-2]: digit 1, followed by a digit from 0 to 2 (for months 10, 11 and 12)And for the year, I used (19|20)\d{2}, which means "19 or 20, followed by 2 digits." That is, every year between 1900 and 2099. That's just an example, you can use \d{4} if you want (remembering that this considers values between 0000 and 9999).Only one detail: the parentheses form one https://www.regular-expressions.info/brackets.html , and when that happens, findall returns a list of tuples with the groups. Using the same text as the previous example:r = re.compile(r'(0[1-9]|[12]\d|3[01]).(0[1-9]|1[0-2]).(19|20)\d{2}') print(r.findall(texto)) The output is:[('01', '02', '20'), ('20', '11', '20'), ('15', '03', '19')That is, a list of tuples, each tuple has a separate day, month and year. In fact, the third element is only the first two digits of the year, because it is the part that is in brackets in regex.To fix this, just change the parentheses to https://www.regular-expressions.info/brackets.html#noncap , putting ?: soon after (. So the regex is like this:r = re.compile(r'(?:0[1-9]|[12]\d|3[01]).(?:0[1-9]|1[0-2]).(?:19|20)\d{2}') print(r.findall(texto)) Now yes the return will be correct:['01.02.2019', '20.11.2018', '15.03.1980']But still, this regex accepts some invalid dates, such as 31.04.2019 (it is invalid, because April only has 30 days), or still 29.02.2019 (invalid because 2019 is not a leap year, so February 2019 has only 28 days). I do not know how your text is and how it is generated, but it may be that you have typing errors, for example, and in such cases it is worth validating the dates.Although it is possible to make a regex to validate all this (including bisexual years verification), it is so complicated that it starts not worth it. https://stackoverflow.com/a/8768241 an example and try to understand it (perhaps as exercise is interesting, but I would never use it in production):^(?:(?:(?:0?[13578]|1[02])(/|-|.)31)\1|(?:(?:0?[1,3-9]|1[0-2])(/|-|.)(?:29|30)\2))(?:(?:1[6-9]|[2-9]\d)?\d{2})$|^(?:0?2(/|-|.)29\3(?:(?:(?:1[6-9]|[2-9]\d)?(?:0[48]|[2468][048]|[13579][26])|(?:(?:16|[2468][048]|[3579][26])00))))$|^(?:(?:0?[1-9])|(?:1[0-2]))(/|-|.)(?:0?[1-9]|1\d|2[0-8])\4(?:(?:1[6-9]|[2-9]\d)?\d{2})$Just to understand this regex will already take a while, imagine to give maintenance. .If you want to validate dates, do away from regexAs we have seen, a simpler regex can bring several excerpts that may not be dates (as we have seen) 00.99.1224 or 31.04.2019). Already a more accurate regex (like the https://stackoverflow.com/a/8768241 already mentioned above) it is so complicated that - in my opinion - it begins not worth using, because it is a maintenance nightmare.Perhaps the best is a half-term: to make a regex not so complicated, not so precise, but that brings something that Look. a date. Then I validate this something to make sure it is indeed a valid date. We can use the regex already seen, which checks days between 1 and 31, and months between 1 and 12, which already serves as a good initial filter.Already to validate the most complicated cases (such as bisexual years and if the month has 30 or 31 days, etc.), we can use the https://docs.python.org/3/library/datetime.html , which possesses the https://docs.python.org/3/library/datetime.html#datetime.datetime.strptime , which transforms a string to date (bassing to this https://docs.python.org/3/library/datetime.html#strftime-strptime-behavior in which this string is), and if the date is invalid, launch a https://docs.python.org/3/library/exceptions.html#ValueError . Then first we can create a function that checks whether a string represents a date in the "day" format. Month. year":from datetime import datetime def data_valida(data): try: datetime.strptime(data, "%d.%m.%Y") return True except ValueError: return False Then we can use this function with the result of findall, and discard the results that are not valid:import re texto = """Data 01.02.2019, outra data 20.11.2018 etc... Outra data 15.03.1980, etc Data inválida: 31.04.2019 """ r = re.compile(r'(?:0[1-9]|[12]\d|3[01]).(?:0[1-9]|1[0-2]).(?:19|20)\d{2}') datas = [data for data in r.findall(texto) if data_valida(data)] print(datas) With this, regex continues to find invalid dates (as 31.04.2019), but the function makes an extra check. In the end, the list will only have the valid dates:['01.02.2019', '20.11.2018', '15.03.1980']Note that, to create the date list, I used the syntax of https://www.pythonforbeginners.com/basics/list-comprehensions-in-python , much more succinct and https://pt.stackoverflow.com/q/192343/112052 . But if you want, you can use one loop more "traditional" and common to other languages:# O loop abaixo é equivalente a: datas = [data for data in r.findall(texto) if data_valida(data)] datas = [] for data in r.findall(texto): if data_valida(data): datas.append(data) print(datas) It still gets better, of course. If your text has something like 112.12.2019 (may be a typing error as it has a 1 the most at first, but it can also be, I don't know, a specific code that by coincidence "seems" with a date), regex will ignore the first 1 and take the rest (12.12.2019).If you want to ignore such cases, we can limit to dates that are "unsolated" in the text. That is, without any other alphanumeric character before or after, and we can do this using the shortcut \b (also called https://docs.python.org/3/library/re.html#index-26 , something like "frontier between words" - https://pt.stackoverflow.com/q/420487/112052 has a more detailed explanation:r = re.compile(r'\b(?:0[1-9]|[12]\d|3[01]).(?:0[1-9]|1[0-2]).(?:19|20)\d{2}\b') With this, cases like 112.12.2019 are ignored by regex.I only made all these suggestions because it was not clear what might be in your text. Depending on how varied your data is, you can adjust the complexity of regex. The important thing is to check if she takes what you need, and - not least - if she doesn't take what you don't need.But as the "don't get what you don't need" part is more complicated, I still find it interesting to validate the dates outside the regex, to make sure they are even dates. But it is at your discretion to do this or not.(1) In Python 3, https://docs.python.org/3/library/re.html#index-27 by \d are any character of http://www.fileformat.info/info/unicode/category/Nd/list.htm , which includes characters such as ٠١٢٣٤٥٦٧٨٩, among others (see https://unix.stackexchange.com/a/414230 for more details, and https://ideone.com/Ax3Wnn to understand better).If you want regex to take only digits from 0 to 9 (ignoring the other characters as ٠١٢٣٤٥٦٧٨٩), you can change \d by [0-9], or still use the https://docs.python.org/3/library/re.html#re.ASCII ( https://ideone.com/IP4Vvm the difference):# para \d não considerar outros caracteres de dígitos, use a flag ASCII r = re.compile(r'\b(?:0[1-9]|[12]\d|3[01]).(?:0[1-9]|1[0-2]).(?:19|20)\d{2}\b', flags = re.ASCII) ou troque \d por [0-9] r = re.compile(r'\b(?:0[1-9]|[12][0-9]|3[01]).(?:0[1-9]|1[0-2]).(?:19|20)[0-9]{2}\b') Of course, if you're working with Portuguese texts, it's a more difficult situation to happen, but in any case, it's the tip.

Note that your method is validaPlaca, therefore it should validate the old plates (which will still be in circulation) and also the new plates in the Mercosur model.So yours regex must validate both formats. An expression to validate the two formats at the same time would be this: [A-Z]{3}[0-9]{1}[A-Z]{1}[0-9]{2}|[A-Z]{3}[0-9]{4}But there are still considerations to do, for example, the plate is of the Mercosur standard but each country https://pt.wikipedia.org/wiki/Placas_de_identifica%C3%A7%C3%A3o_de_ve%C3%ADculos_no_Mercosul .For example:Argentina: AB123CDBrazil: ABC1D23Uruguay: ABC1234Paraguay: 123ABCDIn summary, it would adopt the same regular expression as the Federal Revenue does in the emission of Electronic Fiscal Documents (NF-e, CT-e and MDF-e) that is this: [A-Z]{2,3}[0-9]{4}|[A-Z]{3,4}[0-9]{3}|[A-Z0-9]{7}, in this way the validation of the plates of Brazil would not be restricted.

Just escape the character with \. Just remembering that within a string, the character itself \ https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String#Escape_notation :let word = '\['; let re = RegExp('\\b' + word + '\\b', 'g'); console.log(re); // imprime "/\b[\b/g"This happens because brackets and brackets are https://www.regular-expressions.info/refquick.html in regex. If you want them to "lost their powers" and be interpreted as the characters themselves, you must escape them with \.In the case of the brackets, they define a https://www.regular-expressions.info/charclass.html . For example, [abc] means "the letter a, or the letter b, or the letter c" (only one of them). If a regex has a [, must have the ] corresponding.When you didn't do it escape escape, the result was one [ without the corresponding closure (the character ]) and this makes a mistake, since it is a misformed expression.Using escape escape, it is interpreted as the character itself [, no special meaning (and so you don't need to have the ] corresponding), and there it works.With parentheses it is the same thing:let word = '\\('; let re = RegExp('\\b' + word + '\\b', 'g'); console.log(re); // imprime "/\b\(\b/g"Similar to brackets, brackets https://www.regular-expressions.info/brackets.html in regex: they serve to group a sub-expression and also form a https://www.rexegg.com/regex-capture.html . So one ( must have ) corresponding.But using escape escape, it becomes interpreted as a common character, and therefore does not need the respective closure.O escape escape works for any other character that has special meaning in regex, such as the point (.), which means "any character", the +, which means "one or more occurrences", among others (all cited https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp#Special_characters_meaning_in_regular_expressions ).A detail to look at is that not always the expression without escape escape will give error (as happened with the brackets and brackets), but will eventually generate a different regex. Ex:// usando o "+" sem escape let word = 'a+b'; let re = RegExp('\\b' + word + '\\b', 'g'); console.log(re); // imprime "/\ba+b\b/g" // usando o "+" com escape word = 'a\+b'; re = RegExp('\b' + word + '\b', 'g'); console.log(re); // imprime "/\ba+b\b/g"The first regex is /\ba+b\b/g. In the case, we have a+b, which means "one or more letters a, followed by a letter b" https://regex101.com/r/Ed4tcR/1/ ).The second regex is /\ba+b\b/g. Like now + is escaped, it does not mean "one or more occurrences" and is interpreted only as the character itself +. So the passage a+b means "the letter a, followed by the character +, followed by the letter b" https://regex101.com/r/BymWgI/1/ ).Notice I used it let instead of var to declare the variables. If I didn't know and was curious, https://pt.stackoverflow.com/q/47165/112052 to understand the difference.

You can use this way, validate email, then password using function http://php.net/manual/pt_BR/function.preg-match.php $email = "email@email.com"; $nome = "João Silva"; $regex_nome = '/^[a-zA-ZãÃáÁàÀêÊéÉèÈíÍìÌôÔõÕóÓòÒúÚùÙûÛçǺª\' \']+$/'; $regex_email = '/^[_a-z0-9-]+(\.[_a-z0-9-]+)*@[a-z0-9-]+(\.[a-z0-9-]+)*(\.[a-z]{2,3})$/'; $email = (preg_match($regex_email, $email))?$email:"email inválido"; $nome = (preg_match($regex_nome, $nome))?$nome:"nome inválido"; var_dump($email); var_dump($nome); Can also be used http://php.net/manual/pt_BR/function.filter-var.php for this purpose, give a read.

Yours AppComponent logo on ngOnInit check window.location.pathname which is what the angular router will analyze to find the routes.Remember to enter the main component of your application.if(window.location.pathname.includes('(') || window.location.pathname.includes(')')) { window.location.pathname = window.location.pathname.replace(/\(/g, '%28').replace(/\)/g, '%29'); } Example: https://stackblitz.com/edit/angular-p13och?file=src/app/app.component.ts Another alternative is you put on your index.html a tag script right after the tag closure app-root the content below.<script> window.history.pushState("", "", window.location.pathname.replace(/\(/g, '%28').replace(/\)/g, '%29') + window.location.search); </script> Example: https://stackblitz.com/edit/angular-pyka24?file=src/index.html NOTE: I do not think these solutions are elegant, but I believe it is the only way to solve your problem.

As they said in the other answers, the problem is the passage (?<!,), which is a https://www.regular-expressions.info/lookaround.html#lookbehind . In the case, it checks whether No there is a comma before the desired character (which in case is also a comma). If you do, the regex fails.And then we have (?!,), which is a negative lookahead, which checks if there is no comma later. So... /(?<!,),(?!,)/ serves to capture the commas that do not have another comma before or after, which is another way of saying that regex does not take the cases where there are two or more commas followed ( https://regex101.com/r/NO1jW3/3 ).How are you using this regex in one https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/split , means that the string will be separated only in the positions in which there is a comma (since it does not have a comma before or after). That is, if you have two or more commas in a row, they are not considered in split.Note: At the time when the question was asked, this syntax was not available in all browsers, like Firefox (quoted in the question). But seeing http://kangax.github.io/compat-table/es2016plus/#test-RegExp_Lookbehind_Assertions today - July/2021 - we can see that several other browsers, such as Firefox and Edge, now have support (but anyway, it is not yet something implemented at all, so the alternative below remains an option).Turn your code into Chrome (code below):String.prototype.scapeSplit = function (v) { let r_split = new RegExp('(?<!' + v + ')' + v + '(?!' + v + ')'); let r_replace = new RegExp(v + '{2}'); let s = this.split(r_split); // split produz a lista ["ab", "cd,,ef", "gh,,,ij", "kl"] return s.map(function (x) { return x.replace(r_replace, v); }); } let s = 'ab,cd,,ef,gh,,,ij,kl'; // ["ab", "cd,ef", "gh,,ij", "kl"] console.log(s.scapeSplit(','));How Chrome already supports lookbehindsThe code ran smoothly. I saw that your code first does it split. Using the string 'ab,cd,,ef,gh,,,ij,kl' and doing split with comma, the first regex breaks the string only where there are no two or more commas followed.Then the result is the list ["ab", "cd,,ef", "gh,,,ij", "kl"]. Next is made a map in this list, replacing two followed commas (v + '{2}', which results in ,{2}- two commas followed) by only one. That is, cd,,ef is transformed into cd,ef and gh,,,ij, in gh,,ij.The final result is the list ["ab", "cd,ef", "gh,,ij", "kl"].Alternative to browsers that do not support lookbehindBecause this feature is not supported in all browsers, the approach should be a little different. Instead of split, I will use the method https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/match , and in regex I will use https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp#Parameters , which causes an array to be returned with all matches found.But I will use a different regex, since logic will be inverse. While in split I put a regex with the things that I No I want in the final result (comma that has no other comma before or after), in match I do the opposite: I put the things that I want to be in the final result (in the background, split and match https://www.rexegg.com/regex-style.html#splitvsmatch ). Anyway, what I want is in the final result is:a string that is not commaoptionally followed by a sequence of two or more commasthis whole sequence can be repeated several times (e.g. if you have an excerpt aa,,bb,,,cc,,,dd, all this is a single element split did not separate, so the match must have a regex that considers all this one thing only).In case, I will use ([^,]+(,{2,})?)+. Explaining from the inside out:[^,]+: The delimiter [^ represents a https://www.regular-expressions.info/charclass.html#negated , i.e. regex considers any character other than that between [^ and ]. In the case, you only have the comma. And https://www.regular-expressions.info/repeat.html + means "one or more occurrences." That is, it is a sequence of several characters that are not comma.(,{2,})?: the passage ,{2,} means "two or more commas", and ? makes this whole stretch optional. That means you can have a sequence of several commas or not.O + around all the expression (grouped between parentheses) says that this can be repeated several times. That is, the whole set "various characters that are not comma, followed or not of various commas" can be repeated several times.This ensures that excerpts like ab, ab,,cd and ab,,cd,,,ef will be considered one thing alone. Example:let matches = 'ab,cd,,ef,gh,,,ij,kl'.match(/([^,]+(,{2,})?)+/g); console.log(matches); // ["ab", "cd,,ef", "gh,,,ij", "kl"] The result was the array ["ab", "cd,,ef", "gh,,,ij", "kl"], exactly the same as your original code gets before map. That is, now just do map and your code is ready:String.prototype.scapeSplit = function (v) { let r_match = new RegExp('([^' + v + ']+(' + v + '{2,})?)+', 'g'); let r_replace = new RegExp(v + '{2}'); let s = this.match(r_match); // match produz a lista ["ab", "cd,,ef", "gh,,,ij", "kl"] return s.map(function (x) { return x.replace(r_replace, v); }); } let s = 'ab,cd,,ef,gh,,,ij,kl'; // ["ab", "cd,ef", "gh,,ij", "kl"] console.log(s.scapeSplit(','));The result will be the array ["ab", "cd,ef", "gh,,ij", "kl"].Expressions with more than one characterThe above solution works well when the parameter passed to scapeSplit has only one character.If the parameter has more than one character, there are some modifications to be made.If the browser supports negative lookbehind (as is the case with Chrome), just fix the regex that does the replace to:let r_replace = new RegExp('(' + v + '){2}'); Case v be, for example, the string 12: if you have no parentheses, the result is 12{2} (the number 1, followed by two numbers 2). But I want to actually (12){2} (two occurrences of 12). Fixing this, you can use the string '12' No split that will work smoothly, following the same logic of comma (only separate by 12 if you have no other occurrence 12 before or after).If the browser does not support negative lookbehind, we can't use [^...] as it was done above, so the solution is a bit more complicated1:String.prototype.scapeSplit = function (v) { let r_match = new RegExp('(?:' + v + ')(?!(' + v + ')+)', 'g'); let lookbehind = new RegExp(v + '$'); // simula o lookbehind let indices = [], match; // primeiro obtém os índices em que a expressão ocorre while (match = r_match.exec(this)) { if (match.index == r_match.lastIndex) r_match.lastIndex++; // obtém a substring de zero até o índice em que o match ocorre let leftContext = match.input.substring(0, match.index); if (! lookbehind.exec(leftContext)) { // simular lookbehind negativo indices.push({ start: match.index, end: match.index + match[0].length }); } } // agora faz o split pelas posições encontradas acima let pos = 0; let result = []; indices.forEach(i => { result.push(this.substring(pos, i.start)); pos = i.end; }); // não esquecer do último result.push(this.substring(pos)); let r_replace = new RegExp('(' + v + '){2}'); // o indices.forEach acima produz a lista result = ["ab", "cd1212ef", "gh121212ij", "kl"] return result.map(function (x) { return x.replace(r_replace, v); }); } let s = 'ab12cd1212ef12gh121212ij12kl'; // ["ab", "cd12ef", "gh1212ij", "kl"] console.log(s.scapeSplit('12'));If the parameter is, for example, the string '12', the first regex (r_match) stay (?:12)(?!(12)+). That is, the string 12, provided it is not followed by one or more occurrences of 12.Then I make a while traveling all matches of this regex in the string. Every time I find one, I use Other regex to simulate lookbehind. I do this by getting a substring that corresponds to the original string from the beginning to the point where the match was found (match.index). If this excerpt ends with the string indicated, it means that the lookbehind found a string repeat (but as I want one negative lookbehind, I do if (!lookbehind.exec(leftContext))).For example, if the input string starts with ab12cd, match is found in position 2 (where begins 12). So I make a substring up to position 2 (resulting in ab) and check if this string ends in 12 (i.e. I'm simulating what the lookbehind would do).Then I keep it match.index (position in which match occurred) and match.index + match[0].length (position in which it ends = initial position of match more string size found). At the end of this while, I have all the positions in which the matches occurred. With that I know exactly where I have to do split.Then I make a forEach by these indices, using substring to pick up the indicated excerpt and add these substrings to an array. In the end I just simulated what split I'd do it in case lookbehind be supported.Finally, I do replace to eliminate the repetitions, as was done with the comma (remembering to place the parentheses).PS: the passage if (match.index == r_match.lastIndex) r_match.lastIndex++; is made to fix a bug for cases of zero width matches ( http://www.regexguru.com/2008/04/watch-out-for-zero-length-matches/ ). It does not occur for the specific strings and regex that we are using, but in any case the registry is.(1) - This solution that simulates lookbehind was based https://www.oreilly.com/library/view/regular-expressions-cookbook/9781449327453/ .

No button type="submit", you call the function validar() when the button is clicked (onclick="return validar()"). But then, no onsubmit the form, call the function again. So the first thing is to remove one of these calls. We can remove from button and leave only in form.Then you define several times the function validar. With this, the function is overwritten and https://stackoverflow.com/q/33973987 (in the case, the one validating the name). Here's an example of what happened in your case:function validar() { alert('validar 1'); } // criando a mesma função de novo function validar() { alert('validar 2'); } // e de novo function validar() { alert('validar 3'); } // o alert chamado será "validar 3" validar();When calling the function validar() above, an alert will be displayed with the text "validar 3". This happens because I define the function 3 times, and the third time overwrites the previous ones. That's what happened when you set your function several times. validar().To avoid this problem - and even organize the code - you must change the names, indicating what each function validates (e.g. validarNome, validarCPF, etc), and then have a function that validates everything, that is, that calls all these functions.Another detail is that you do not need to call submit() within these functions. Putting validation call in onsubmit the form, just return true or false, https://stackoverflow.com/q/35037069 .In the example below I gave a "ready" in your HTML (retreat the CSS classes, etc.), just to focus on the validation mechanism itself.function formatar(mascara, documento) { let i = documento.value.length; let saida = mascara.substring(0, 1); let texto = mascara.substring(i); if (texto.substring(0, 1) != saida) { documento.value += texto.substring(0, 1); } } function validarDtNasc() { let value = document.getElementById("nascimento").value; let re = /^[0-9]{2}/[0-9]{2}/[0-9]{4}$/; if (!re.test(value)) { // campo inválido, retorna false para o formulário não ser submetido alert('Data de Nascimento Inválida'); document.form.nascimento.focus(); return false; } return true; } function validarCPF() { let value = document.getElementById("cpf").value; let re = /([0-9]{2}[.]?[0-9]{3}[.]?[0-9]{3}[/]?[0-9]{4}[-]?[0-9]{2})|([0-9]{3}[.]?[0-9]{3}[.]?[0-9]{3}[-]?[0-9]{2})/g; if (! re.test(value)) { // campo inválido, retorna false para o formulário não ser submetido alert('CPF Inválido'); document.form.cpf.focus(); return false; } return true; } function validarTel() { let value = document.getElementById("tel").value; let re = /^[0-9]{2}-[0-9]{5}-[0-9]{4}/g; if (! re.test(value)) { // campo inválido, retorna false para o formulário não ser submetido alert('Telefone Inválido'); document.form.tel.focus(); return false; } return true; } function validarNome() { let value = document.getElementById("nome").value; let re = /^[a-zA-ZéúíóáÉÚÍÓÁèùìòàçÇÈÙÌÒÀõãñÕÃÑêûîôâÊÛÎÔÂëÿüïöäËYÜÏÖÄ-\ \s]+$/; if (!re.test(value)) { // campo inválido, retorna false para o formulário não ser submetido alert('Nome Inválido'); document.form.nome.focus(); return false; } return true; } // valida todos os campos function validarTudo() { // se um deles for inválido, retorna false e o form não é submetido return validarNome() && validarTel() && validarDtNasc() && validarCPF(); }<form name="form" id="form" onsubmit="return validarTudo();"> <div> <label for="nome">Nome:</label> <input type="text" name="nome" id="nome" required autocomplete="off"> </div> <div> <label for="tel">Tel:</label> <input type="text" name="tel" id="tel" required autocomplete="off" maxlength="13" minlength="13" onkeypress="formatar('##-#####-####', this)"> </div> <div> <label for="email">Email:</label> <input type="email" name="email" id="email" required autocomplete="off"> </div> <div> <label for="nascimento">Data Nascimento</label> <input type="text" name="nascimento" id="nascimento" required autocomplete="off" maxlength="10" minlength="10" onkeypress="formatar('##/##/####', this)"> </div> <div> <label for="cpf">CPF:</label> <input type="text" name="cpf" id="cpf" autocomplete="off" required maxlength="14" minlength="14" onkeypress="formatar('###.###.###-##', this)"> </div> <div> <button type="submit">Confirmar</button> </div> </form>Also note that to declare the variables changed var by let - https://pt.stackoverflow.com/q/47165/112052 between them.About validationsAs for the validations itself, it is important to highlight some points.The regex of the name accepts several spaces. Fill the field with only spaces and you will see that it considers valid. It also accepts names without last name (write only "John" is valid, it is up to you to check whether this is actually valid or not on your system) Maybe https://regex101.com/r/H9m60i/1 be more accurate, assess whether you need something so complex or leave to make a more accurate validation in https://pt.stackoverflow.com/q/13298/112052 .In the CPF, it is interesting to also validate the verifier digits, take a look https://pt.stackoverflow.com/q/51340/112052 for more details (and do some searches on the site and on the internet in general, what is not missing are examples for you to base yourself).About regular expressions, you can use \d, which is synonymous with [0-9], as described in https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Regular_Expressions#Using_special_characters . That is:// em vez dessa let re = /^[0-9]{2}/[0-9]{2}/[0-9]{4}$/; // pode usar essa let re = /^\d{2}/\d{2}/\d{4}$/; About the use of regex to validate dates, I comment a little https://pt.stackoverflow.com/a/363711/112052 and in the second half https://pt.stackoverflow.com/a/360903/112052 , and there are also examples https://pt.stackoverflow.com/q/272539/112052 . But perhaps it is better to follow the recommendation https://pt.stackoverflow.com/a/272541/112052 and instead of regex, use the language resources that are specific to treat dates, such as the object itself https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Date (There are also https://pt.stackoverflow.com/search?q=javascript%20Date on the site) or even a specific lib, such as https://momentjs.com/docs/#/parsing/is-valid/ .And if you want, you can also change the field to input type="date", not forgetting to check the https://caniuse.com/#feat=input-datetime (and documentation https://developer.mozilla.org/en-US/docs/Web/HTML/Element/input/date#Handling_browser_support for browsers that do not support this type of input).And don't forget too https://pt.stackoverflow.com/q/13298/112052 .

First of all, you don't need two pairs of parentheses in a row.When there are parentheses, they form one https://www.regular-expressions.info/brackets.html . That is, what is inside them will be available in one https://docs.python.org/3/library/re.html#re.Match.group , if regex finds a match. This is the standard behavior of the parentheses.But when you don't want to capture a group, and just group sub-expressions, you use syntax (?:, defining a https://www.regular-expressions.info/brackets.html#noncap . Basically, this says that the parentheses should not create a group if regex finds a match.So do ((?:...)) It's kind of weird. It is a non-capture group within a catch group, that is, you said you want to capture something, and that something is "a thing I don't want to capture". Contradictory and unnecessary, then the first thing is to eliminate this (and depending on the case, choose only one of them if it is accurate).Anyway, what's inside the parentheses is exmo|excelentissimo, which means "the string exmo or a string excelentissimo". This regex takes only this stretch, ignoring the remainder of the string.To get the whole line, just use a regex that takes all the text:import re text = """ TEXTO ACIMA EXMO. SR. DR. JUÍZ DE DIREITO DO JUIZADO ESPECIAL CÍVEL DA COMARCA DA CAPITAL TEXTO ABAIXO EXCELENTISSIMO. SR. DR. JUÍZ DE DIREITO DA VARA ESPECIAL CÍVEL DA COMARCA DA CAPITAL MAIS TEXTO""" results = re.findall(r'^(?:EXMO|EXCELENTISSIMO). SR. DR. JUÍZ DE DIREITO (?:DO JUIZADO|DA VARA) ESPECIAL CÍVEL DA COMARCA DA CAPITAL$', text, re.IGNORECASE | re.MULTILINE) for result in results: print(result) Notice I used it (?:EXMO|EXCELENTISSIMO) and (?:DO JUIZADO|DA VARA) for parts that may be one thing or another. I used non-capture groups - notice syntax (?: - Yeah. https://docs.python.org/3/library/re.html#re.findall , when there are capture groups in regex, only groups are returned. But as you want the entire string, I used non-capture groups.The rest of the text, as it does not change, can be exactly what you want ("SR. DR. JUIZ etc"). The only exception is the point, since in regex it has special meaning (meaning https://www.regular-expressions.info/dot.html ), then it must be escaped and written as ..Other details are https://www.regular-expressions.info/anchors.html ^ and $, which usually mean the beginning and end of the string. But thanks to https://docs.python.org/3/library/re.html#re.MULTILINE , they are also interpreted as the beginning and end of a line. With this you guarantee that the line contains only what is specified in regex.The above code prints:EXMO. SR. DR. JUÍZ DE DIREITO DO JUIZADO ESPECIAL CÍVEL DA COMARCA DA CAPITAL EXCELENTISSIMO. SR. DR. JUÍZ DE DIREITO DA VARA ESPECIAL CÍVEL DA COMARCA DA CAPITAL How you used it https://docs.python.org/3/library/re.html#re.IGNORECASE , regex will also pick up stretches with tiny letters ("exmo. sr. dr. etc....") and even small and mixed capitals ("eXMo. SR. Dr...). If you want it to take only capital letters, remove this flag (but don't forget to keep it MULTILINEresults = re.findall(r'^(?:EXMO|EXCELENTISSIMO). SR. DR. JUÍZ DE DIREITO (?:DO JUIZADO|DA VARA) ESPECIAL CÍVEL DA COMARCA DA CAPITAL$', text, re.MULTILINE) When I was editing your question, I saw that the text was occupying 3 lines:EXMO. SR. DR. JUÍZ DE DIREITO DO JUIZADO ESPECIAL CÍVEL DA COMARCA DA CAPITAL I don't know if it's like that or if it should be all on a line. In any case, we can exchange space after "DIREITO" and "ESPECIAL" per \s, which considers both space when line break:results = re.findall(r'^(?:EXMO|EXCELENTISSIMO). SR. DR. JUÍZ DE DIREITO\s(?:DO JUIZADO|DA VARA) ESPECIAL\sCÍVEL DA COMARCA DA CAPITAL$', text, re.MULTILINE) If you want, you can also use [ \n] (a space or \n - notice there's a space after [), since \s also considers other characters, such as _ and others mentioned in https://docs.python.org/3/library/re.html#regular-expression-syntax .

In the question it is mentioned that you want to take only the first number, but then it says that you also want to catch everyone. So I left a solution for every case.Take only the first numberIf you just want to extract an excerpt from the string that corresponds to regex, you can use http://php.net/manual/en/function.preg-match.php , passing your string and a variable that will contain the matches:$string = 'QD 10 LT 20 PA 30'; preg_match('/^\D+(\d+)\D+/', $string, $matches); echo($matches[1]); In case, I used the https://www.regular-expressions.info/anchors.html ^, which indicates the beginning of the string. Then I check up to the first digit group ((\d+)). As you just want to know from the first group, you don't need to check the remainder of the string.The variable $matches will contain the passages that regex captures. And how \d+ is within parentheses, a https://www.regular-expressions.info/brackets.html . And as is the first pair of parentheses, it will be available in $matches[1]. That's why this code prints 10.There's a detail in the regex. Soon after ^ there is \D+ (one or more characters other than numbers). This means that the string cannot start with numbers. That is, this code does not print anything, because as the string starts with numbers, the preg_match returns false:$string = '11 QD 10 LT 20 PA 30'; // começa com números (não entra no if) if (preg_match('/^\D+(\d+)\D+/', $string, $matches)) { echo ($matches[1]); } If so, an alternative is to change the https://www.regular-expressions.info/repeat.html + (one or more occurrences) by * (zero or more occurrences):$string = '11 QD 10 LT 20 PA 30'; // começa com números if (preg_match('/^\D*(\d+)\D+/', $string, $matches)) { // troquei o primeiro \D+ por \D* echo ($matches[1]); // agora entra no if } But there is another. From what I understand, you also want to validate if the string is in the format "two letters, two numbers, two letters, two numbers, etc."Let's assume that the format is exactly that (always two letters and two numbers, separated by space, and this pattern can be repeated several times). In this case, we can change the + by {2}, which means https://www.regular-expressions.info/repeat.html#limit :$string = 'QD 10 LT 20 PA 30 SC 40'; if (preg_match('/^[A-Z]{2} (\d{2})( [A-Z]{2} \d{2})*$/', $string, $matches)) { echo ($matches[1]); // 10 } First we have the beginning of the string (^), and then [A-Z]{2} (\d{2}):[A-Z]{2} means "2 letters of A a Z" (only capital). If you want only tiny letters, change by [a-z]{2}, and if both capitalizes and lowercases, use [A-Za-z]{2}then we have a blank space (since the fields are separated by a space)we have (\d{2}): two digits within a catch group (so that it is in $matches[1])Then we have ( [A-Z]{2} \d{2})*. Within the parentheses we have a space (repare in space right after the (), followed by two letters, another space and two digits. Outside the parentheses we have *, which means that all this sequence (space, 2 letters, space, 2 digits) can repeat zero or more times. This ensures that your string can be so much QD 10 LT 20 PA 30 SC 40 how QD 10.Finally, we have $, which means "end of string". With this I guarantee that the entire string has the format I need, and still catch the first digit group in $matches[1].If the amount of letters and digits can vary, just change the quantifier. Examples:\d{1,5} - between 1 and 5 digits\d{2,} - 2 or more digits (no maximum limit)Adjust to the values you need.Get all numbersIn this case, use http://php.net/manual/en/function.preg-match-all.php :$string = 'QD 10 LT 20 PA 30 SC 40'; if (preg_match_all('/\b(\d{2})\b/', $string, $matches, PREG_SET_ORDER)) { foreach ($matches as $m) { echo $m[1] . "\n"; } } In this case, I'm using \b as a delimiter, because it represents a https://www.regular-expressions.info/wordboundaries.html , which in practice checks whether there is any non-alpha-numeric character before and after (\d{2}). The difference is now $matches will contain an array with all 2-digit occurrences of the string. The output is:10203040The difference is that preg_match_all is only taking the digits, but does not actually validate the string format. That is, if the string is abcdef 10, xyz!20, yet she will get the numbers.But you can use the previous regex only to validate the format, and then use preg_match_all to extract the numbers:$string = 'QD 10 LT 20 PA 30 SC 40'; // verifica o formato da string if (preg_match('/^[A-Z]{2} (\d{2})( [A-Z]{2} \d{2})*$/', $string)) { // só extraio os números se o formato estiver OK if (preg_match_all('/\b(\d{2})\b/', $string, $matches, PREG_SET_ORDER)) { foreach ($matches as $m) { echo $m[1] . "\n"; } } }

Put your WHERE as follows:WHERE DATEDIFF(STR_TO_DATE(LEFT(data_gabarito,10),'%d/%m/%Y'), NOW()) >= -2

When you do a query like this:select campo from tabela where campo REGEXP '[0-9]{2}/[0-9]{2}/[0-9]{4}'; You are bringing the field (in case all the value 30/09/2018 e 14/10/2018 e 21/10/2018), provided it satisfies regex (i.e. as long as it has numbers in the given format). That's why query brings "everything."To extract only one part of the field, use https://dev.mysql.com/doc/refman/8.0/en/regexp.html#function_regexp-substr (available from MySQL 8). Supposing the table is:create table tabela ( campo varchar(200) ); insert into tabela(campo) values('30/09/2018 e 14/10/2018 e 21/10/2018'); insert into tabela(campo) values('blablabla e 14/10/2018 e 21/10/2018'); insert into tabela(campo) values('nao tenho nenhuma data'); We can do:select REGEXP_SUBSTR(campo, '[0-9]{2}/[0-9]{2}/[0-9]{4}') from tabela; And the result will be:30/09/2018 14/10/2018 null The third field returns null, since the value of the field does not correspond to the regex. You can change the query to only return the desired snippet when the field has a date in the desired format:select REGEXP_SUBSTR(campo, '[0-9]{2}/[0-9]{2}/[0-9]{4}') from tabela where campo REGEXP '[0-9]{2}/[0-9]{2}/[0-9]{4}'; Now where check if the field has any date, and REGEXP_SUBSTR extract only the desired excerpt. The result is:30/09/2018 14/10/2018 If you want the second occurrence, just pass 2 additional parameters: the initial position and the occurrence:--- traz a segunda ocorrencia select REGEXP_SUBSTR(campo, '[0-9]{2}/[0-9]{2}/[0-9]{4}', 1, 2) from tabela where campo REGEXP '[0-9]{2}/[0-9]{2}/[0-9]{4}'; In the case, the initial position is 1, and the occurrence is 2 (i.e., I will take the second date present in the field). The result is:14/10/2018 21/10/2018 See this example running on https://www.db-fiddle.com/f/xrj7Raj4bJU6edizXumAsb/0 .But there is one: if you have something like that 112/10/2019, the regex will take the stretch 12/10/2019. To avoid these cases, you can change the regex to:(^|[^0-9])[0-9]{2}/[0-9]{2}/[0-9]{4}([^0-9]|$) At first there is the passage (^|[^0-9]): | https://www.regular-expressions.info/alternation.html . Therefore, this passage means https://www.regular-expressions.info/anchors.html (^) or "anything other than number" [^ means https://www.regular-expressions.info/charclass.html#negated , that is, I don't want 0-9 - no digits from zero to 9).And in the end there is ([^0-9]|$): any character other than number or the end of the string ($). This ensures you won't take cases like 112/10/2019. https://www.db-fiddle.com/f/ravpd8SN1wXRnrzTVBdYdQ/1 the difference.In other languages and engines it is possible to use \b instead of (^|[^0-9]) and ([^0-9]|$)But MySQL can't stand it.Before MySQL 8 there was REGEXP_SUBSTR, so it's a little more complicated. If the date is right at the beginning of the string, you can use the https://pt.stackoverflow.com/a/360864/112052 . But if the date is in any position of the string, I find it easier to extract the data using some programming language and manipulate them there (because languages usually have more regex features than MySQL, which is https://www.regular-expressions.info/mysql.html ).Dates x RegexOnly one Addendum on the use of regex to check dates. A regex [0-9]{2}/[0-9]{2}/[0-9]{4} accepts values that are not valid dates, as 99/99/9999 and 00/00/0000, https://regex101.com/r/VnMrWl/1 . That's because [0-9] means "any digit 0 a 9".It is possible to improve a bit by using something like:(0[1-9]|[12][0-9]|3[01])/(0[1-9]|1[012])/(19|20)[0-9]{2} This ensures that the days can only be from 01 to 31, the months from 01 to 12, and the years from 1900 to 2099.But still yes, this regex accepts dates like 31/04/2019 (April only has 30 days) and 29/02/2019 (February 29 in a non-bisexual year), https://regex101.com/r/VnMrWl/2 . It was not clear how the data is inserted into your database, but it may be that a typing error occurs, or simply an invalid date appears in the middle of the string.You can even have a regex that checks these conditions. https://stackoverflow.com/q/15491894 there are some examples, but they are so complicated that it begins not worth it. See one of them, just to get an idea:^(?:(?:31(/|-|.)(?:0?[13578]|1[02]))\1|(?:(?:29|30)(/|-|.)(?:0?[1,3-9]|1[0-2])\2))(?:(?:1[6-9]|[2-9]\d)?\d{2})$|^(?:29(/|-|.)0?2\3(?:(?:(?:1[6-9]|[2-9]\d)?(?:0[48]|[2468][048]|[13579][26])|(?:(?:16|[2468][048]|[3579][26])00))))$|^(?:0?[1-9]|1\d|2[0-8])(/|-|.)(?:(?:0?[1-9])|(?:1[0-2]))\4(?:(?:1[6-9]|[2-9]\d)?\d{2})$Do you understand? Me neither.Remembering \d is not supported by MySQL, so you should trade them by [0-9]. Even so, in my opinion it is not worth using this regex. It is a problem to understand and to keep in the future (good luck for those who have to change it and add new cases, for example).You have to find a balance. Using the giant regex above ensures that no invalid dates will be returned, but it is a maintenance nightmare. Using the suggested expressions at the beginning of this response, you get excerpts that look. a date (can hit most cases, but it will also bring some invalid dates, something you should evaluate if it is acceptable). Once extracted from these excerpts, additional validation could be done later, preferably using some date API (i.e. extract these database data and validate using your favorite programming language).Regex is a powerful tool, but https://blog.codinghorror.com/regular-expressions-now-you-have-two-problems/ .

O s/ and //g are not part of the regex itself. This syntax is used in other languages (and in some commands, such as sed), but in Java you only need to pass the regular expression as parameter:replaceAll("<!--(.|\\s)*?-->", "") However, this expression is not very good in terms of efficiency, and can even give StackOverflowError https://stackoverflow.com/a/7510006 (as https://pt.stackoverflow.com/questions/360486/remover-tag-de-coment%C3%A1rio-html-com-java#comment720338_360486 ).Basically, the https://www.regular-expressions.info/alternation.html (indicated by |) makes the two alternatives checked if necessary. Most often it will fall into the first case (the point, which means "any character", but that by default actually means "any character except https://www.regular-expressions.info/dot.html "). Every time regex finds a line break, the alternation will test the first option and then the second. In addition, the lazy operator (*?), although convenient for your case, https://www.rexegg.com/regex-quantifiers.html#lazy_expensive . For small strings, these details don't matter, but for big strings it starts to make a difference (and I'm assuming that your file is big enough for it to make a difference, since you mentioned gave it StackOverflowError).You can remove the alternation if the point also matches line breaks, using the option https://docs.oracle.com/javase/8/docs/api/java/util/regex/Pattern.html#DOTALL . In the case of the method replaceAll, it is not possible to pass this option as parameter, but you can enable it by placing (?s) at the beginning of regex:arquivo.replaceAll("(?s)<!--.*?-->", ""); This should already improve a little regex performance. I made a comparison at regex101. and see that https://regex101.com/r/dapW65/4/debugger need more steps to verify the string, compared to the https://regex101.com/r/qx0YPe/5/debugger .Of course. that these numbers will vary according to strings, since each language has a engine with own implementation details, some optimize some cases, etc. But in general, removing alternation already greatly increases performance. How I don't have your full file and couldn't simulate it StackOverflowError, I'm based on the regex101 tests. with (but I suggest you do tests with the files in fact to be sure).The other advantage is for cases of having poorly formed comments (speaking the closing tag, for example). See that https://regex101.com/r/dapW65/6/debugger needs more than 2700 steps to realize that the comment has no closure, while the https://regex101.com/r/qx0YPe/7/debugger needs only 140. Even if your file does not have this, the general performance of the second version, compared to the first, already justifies the change.AlternativeYou can still optimize a bit more by using the regex below:arquivo.replaceAll("<!--(?>[^-<>]*)(?>(?!-->)[-<>][^-<>]*)*-->", ""); She uses a technique known as https://www.perlmonks.org/?node_id=89547 , and is described in more detail http://shop.oreilly.com/product/9780596528126.do . But basically it consists of identifying 3 basic elements of the stretch you want to capture:the delimiters: in our case, are <!-- and -->, which appear at the beginning and end of regexthe "normal": that is, what is most frequent that appears among the delimiters. In case, I used [^-<>]* (zero or more characters other than hyphen, < or >)o "special": the character that is not normal (most frequent in a comment) and/or that may mean that we find the final delimiter (in the case, I used [-<>] - a hyphen, or < or >)The general format of regex is delim normal* (especial normal*)* delim. Also used https://www.regular-expressions.info/atomic.html (indicated by (?>)), which make the engine not do backtracking (this happens when she doesn't find one match, but back a few steps to try other string combinations, making it take longer). But as the "normal" and "special" are mutually exclusive, the backtracking will be done out, so the atomic group skips these unnecessary steps.I also use a https://www.regular-expressions.info/lookaround.html (The passage (?!-->)). Basically, this passage checks if something does not exist ahead (in case I check if there is no comment closure -->). If you don't, the regex goes on [-<>] (hyphen or < or >), followed by zero or more characters that are not hyphens (and all this stretch can be repeated several times, because there is another * outside the parentheses). This ensures that hyphens and even < or > inside the comment (e.g. <!-- <!-- abc ->-----> is a single comment). https://regex101.com/r/40DzX6/7/debugger that performance improves slightly compared to the second version. Again, for small strings, the difference will be irrelevant, but for large strings - as it seems to be your case - can make a big difference.Use an HTML parserBut maybe regex https://stackoverflow.com/a/1732454 for your case. Why not use an HTML parser?I made the example below with the https://jsoup.org/ Version 1.8.3:import org.jsoup.Jsoup; import org.jsoup.nodes.Document; import org.jsoup.nodes.Node; public void removerComentarios(Node node) { for (int i = 0; i < node.childNodeSize();) { Node child = node.childNode(i); if (child.nodeName().equals("#comment")) { child.remove(); } else { removerComentarios(child); i++; } } } Document doc = Jsoup.parse(arquivo); removerComentarios(doc); System.out.println(doc.html()); Of course, there is still the problem of the file being too large and can burst the memory, but it still seems to me to be a simpler solution than a regex. If jsoup doesn't answer you, you can choose another https://en.wikipedia.org/wiki/Comparison_of_HTML_parsers .

The brackets define a https://www.regular-expressions.info/charclass.html . When you do [A-Za-z], that means "a letter of a a z or of A a Z". That is, it can be both a tiny and capital letter.So if you have a capital, but no lowercase (or vice versa), this regex lets you pass. To make you have at least one of each (you have to have a capital and a tiny), and also make sure you also have a number, and that you have at least 6 characters, you have 2 alternatives.1- check each condition separatelyFor each condition, we can make one preg_match different. The detail is the criterion "must have at least 6 characters". We already know which letters and numbers are required, but what characters can the password have? Will you allow special characters? Accents? Emojis? And that way. The more options, https://github.com/mathiasbynens/emoji-regex/blob/v7.0.3/es2015/index.js#L5 . Just to simplify, let's assume that the password can have letters, numbers, _, @ and $. That's just for exemplifying, since limiting too much allowed characters decreases the amount of possible passwords, but allowing too much can lead to cases where the user will have difficulties typing the password again, as discussed https://security.stackexchange.com/q/2096 .Finally, for letters, numbers and _ we can use https://www.regular-expressions.info/shorthand.html \w. Now @ and $, we can use the characters themselves, so the expression is [\w$@] (again, this is just to simplify the example – add all the characters you need inside the brackets). And for size, we use {6,}, which means "6 or more occurrences". That is, you must have at least 6 characters. The code would look like this:function senhaValida($senha) { return preg_match('/[a-z]/', $senha) // tem pelo menos uma letra minúscula && preg_match('/[A-Z]/', $senha) // tem pelo menos uma letra maiúscula && preg_match('/[0-9]/', $senha) // tem pelo menos um número && preg_match('/^[\w$@]{6,}$/', $senha); // tem 6 ou mais caracteres } var_dump(senhaValida('aB1@xy$z')); // true var_dump(senhaValida('aB1')); // false, não tem 6 caracteres var_dump(senhaValida('AB1@XYZ')); // false, não tem letra minúscula var_dump(senhaValida('ab1@xyz')); // false, não tem letra maiúscula var_dump(senhaValida('ABc@xyz')); // false, não tem número Notice that for size I also used the https://www.regular-expressions.info/anchors.html ^ and $, which respectively mean the beginning and end of the string. With this I guarantee that, from start to finish, there are at least 6 occurrences of the characters I set (and only those characters).Other conditions do not need ^ and $, because I am only checking if the characters exist in any position of the string.The code prints bool(true) for the first password, and bool(false) for the rest.2- do everything in a regex onlyFor that, we have to use https://www.regular-expressions.info/lookaround.html . The idea of lookahead is to check an expression, and then go back where it was and proceed with the rest of the regex. She's like this:function senhaValida($senha) { return preg_match('/^(?=.[a-z])(?=.[A-Z])(?=.[0-9])[\w$@]{6,}$/', $senha); } First we have ^ (the beginning of the string). Then we have one lookahead, which begins with (?=. In it we have .[a-z], which means "zero or more occurrences of any character" (.*), followed by a tiny letter ([a-z]). That is, this lookahead verifies if there is a tiny letter in the string.The trick of lookahead is that, after it checks if the tiny letter exists, it goes back to where it was (i.e. to the beginning of the string) and checks the remainder of the expression. And the rest, we have another lookahead to check if there is a capital letter, and then another one to check if there is a number. If any lookaheads fail, regex as a whole also fails.And then we have the expression that checks if there are at least 6 characters, followed by the end of the string ($). That is, it does all the checks of the previous example at once.

Also available in https://ideone.com/GrH0ZF .Try this:using System.Text.RegularExpressions; using System.Collections.Generic; using System; using System.Linq; public class Program { public static void Main(string[] args) { Regex r = new Regex(@"[\d-]+"); var matches = r.Matches("9000-001 a 9999-999"); foreach (var match in matches.Cast&lt;Match&gt;()) { Console.WriteLine($"Encontrada a seguinte ocorrência: '{match.Value}'"); } } } where, in the regular expression:\d is the wildcard+ is the quantifier for one or more occurrencesSee what type occurrences 999- or -12409- will also be obtained with the regular expression provided. To circumvent this and keep the occurrence format shown, use the regular expression \d+-\d+.

Understanding:Start with a letter: a-z;You must start with a letter a a z tiny. OK!Contain only letters, number and characters ,, ., - and _;How does not inform a minimum number of characters after the first character that is required ("You must start with a letter..."), it is considered (by what I understood) that from the second character onward is optional; but if you have, you must obey the above criteria.Logo:Within the criteria exposed in the question, you could use this expression:^[a-z][a-z\d_,\-.]*$ Explanation:^ Delimita o início da string. [a-z] O primeiro caractere deve ser de "a" a "z" (minúsculas). [a-z\d_,\-.] Segundo caractere em diante. A partir do segundo caractere, pode ou não ter: [a-z] letras de "a" a "z" (minúsculas) \d números _ underline , vírgula \- hífen (escapado) . ponto * Quantificador que irá encontrar nada ou quantas ocorrências tiver a partir do segundo caractere. $ Delimita o fim da string. Example using a function:<?php function checa($str){ return preg_match('/^[a-z][a-z\d_,-.]*$/', $str); } // 0 = false, 1 = true echo checa("a"); // 1 echo checa("1"); // 0 echo checa("h1-11,11,"); // 1 echo checa("Aa"); // 0 echo checa("1abc_"); // 0 echo checa("aaaaaaaa"); // 1 echo checa("a-----,"); // 1 echo checa("a-12,_#,"); // 0 echo checa(""); // 0 ?> To check, you can use one if:<?php if(checa("a")){ // com o valor "a" irá entrar aqui. // significa que a string começa com uma letra minúscula // e atende ao critério da regex echo "passou"; }else{ echo "não passou"; } ?> Remembering that upper and sharp letters will be considered invalid.

If they are the users themselves who put the phones in the comments, then there is not much control over the format.Of course you can consider some more common formats. By their examples, I saw that it can be the 9 digits of the mobile number, all together or separated by spaces (999999999, 9 9999 9999 or 99999 9999), being the optional DDD.A https://pt.stackoverflow.com/a/357752/112052 suggests a very complex regex to contemplate these (and many others) cases, but the https://dev.mysql.com/doc/refman/8.0/en/regexp.html#regexp-syntax Unfortunately https://www.regular-expressions.info/mysql.html and will not support all proposed resources, as https://www.regular-expressions.info/shorthand.html or https://stackoverflow.com/a/39727000 (excerpts starting with (?=, (?! and (?<!).Thus, it follows an alternative that checks some formats:select * from tabela where comentario REGEXP '(^|[^0-9])(\\(?0?[0-9]{2}\\)?)?9 ?[0-9]{4} ?[0-9]{4}([^0-9]|$)'; (^|[^0-9]): | https://www.regular-expressions.info/alternation.html . Therefore, this passage means " https://www.regular-expressions.info/anchors.html "^) or "anything that is not number" (the [^ means https://www.regular-expressions.info/charclass.html#negated , that is, I don't want 0-9 - no digits from zero to 9).This ensures that so far I can be at the beginning of the string (it will be that the phone is already at the beginning), or it has any character that is not a number (avoiding you take cases like 3393333333333333333).Then we have (\\(?0?[0-9]{2}\\)?)?. We go through parts, from inside out:0?[0-9]{2} - an optional zero (0? - ? indicates " https://www.regular-expressions.info/repeat.html ", what is the same as saying "optional"), followed by 2 digits ([0-9] is any digit from 0 to 9, and {2} what to say " https://www.regular-expressions.info/repeat.html#limit "), because the DDD can be written as 11 or 011\\(? and \\)? - the parentheses can be optional. I did so because only the parentheses ( and ) have special meaning in regex, as they serve https://www.regular-expressions.info/brackets.html . That's why we have to escape them with \\.Finally, all this passage is in brackets (i.e. grouped into a single sub-expression), and ? in the end makes this whole stretch optional.That is, the DDD is optional.Then we have 9 ?, which is number 9 followed by an optional space (repare that there is a space before the ?, that is, the space that is optional, not the 9). Here I am assuming that it will be only cell numbers that start with 9 - remembering that in the future we will be able to have phones that start with 8, 7, etc, so it is at your discretion to always leave 9 or switch to [0-9] (or) [7-9] if you want to start only with 7, 8 or 9, etc.).Then we have [0-9]{4} (4 digits), followed by an optional space, plus 4 digits.And finally, we have ([^0-9]|$): any character other than number or the end of the string ($). This also ensures that you won't get more digits than necessary, preventing you from taking for example 3393333333333333333. http://sqlfiddle.com/#!9/9a4e837/8 you can see this query working.If you want to increment, you can place the separator as a hyphen or space, for example, such numbers would be accepted as 9 9123-4567 or 99123-4567. Just change the optional spaces per [ \\-]? (a space or a hyphen, optional). The regex would look like this:select * from tabela where comentario REGEXP '(^|[^0-9])(\\(?0?[0-9]{2}\\)?)?9[ \\-]?[0-9]{4}[ \\-]?[0-9]{4}([^0-9]|$)'; http://sqlfiddle.com/#!9/85a631/2 it working.You can also add an optional space after DDD:(^|[^0-9])(\\(?0?[0-9]{2}\\)?)? ?9 ?[0-9]{4} ?[0-9]{4}([^0-9]|$) ^^ Without this space, the DDD is ignored for cases like (11) 9 9123 4567 - only the phone number is captured by regex, but the DDD does not, https://regex101.com/r/Qpi5bL/1 . Already placing the optional space, the DDD is also captured, https://regex101.com/r/Qpi5bL/2 .Another detail is that we are considering only mobile numbers. But there are still 8-digit phone numbers (in residences not so much, but in companies, it is still quite common). If you also want to consider these numbers, just put 9 as optional:(^|[^0-9])(\\(?0?[0-9]{2}\\)?)? ?9? ?[0-9]{4} ?[0-9]{4}([^0-9]|$) ^^ Just remembering that, as are the users who type their numbers, there may always be some strange format that you did not predict. And the more possibilities, the more complex the regex gets.For example, the regex I suggested only considers an optional blank space. But if you want me to have more than one space, you can change ? by * (zero or more occurrences), or limit the amount with keys (e.g. {0,3} limits between 0 to 3 occurrences).There is also the possibility of a CPF being confused with a phone, since both can be written without any separator (43912341222 can be both a CPF and a DDD + phone - even if people custom write the CPF as 439.123.412-22, who guarantees you won't have a case like that? Anyway, see if this applies to your cases). Anyway, regex isn't such a "magic" business, and it's up to you to assess whether you're going to trust her so much to the point of automatically removing anything she takes...Perhaps it is better to follow the suggestion given in https://pt.stackoverflow.com/a/357752/112052 and verify this before of inserting into the bank. I don't know what language you're using, but https://www.regular-expressions.info/tools.html , which allows writing expressions https://regex101.com/r/PN7pJk/1 , for example (using \b to delimit the phones, without having to use the "truck" I did above with (^|[^0-9]) and ([^0-9]|$), besides use \d as a shortcut to [0-9] and \s for spaces).