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Establish regular expression

Z
zymir last edited by

Hello, such a problem, it is necessary to write a regular expression that would give everything that is between the words =begin and =end. If the case is - =begin ... =end ... =end, then only that goes back to the first =end. I wrote it myself, "=begin(.*)=end, but it only works when the text fits in one line.
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Nykeriab

using System.Text.RegularExpressions;
// ...
var str = "=begin 1 =end 2 =end";
var m = Regex.Match(str, "=begin(.+?)=end", RegexOptions.Singleline);
m.Groups[1].Value      // " 1 "

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Expression https://www.regexplanet.com/share/index.html?share=yyyyyr6du4d Only a line that consists of one symbol (primary letter, {, lap, numbers, }Because the template is a non-quantified symbolic class with the symbolic classes in place (in the so-called character class union, i.e. between these symbolic classes of OR relationship), and the cuntifiers are recognized as a letter text ([{6}] I will. {♪ 6 or })You need to use it.^[А-Я] \d{6}$ https://regex101.com/r/Pr0Kxm/1/ If "problem" means any Laundry symbolto replace the gap in terms \s and use the flag. (?U) (when all the Junicoda gaps are found):(?U)^[А-Я]\s\d{6}$ Details^ - starting line (in method) matches() You can put it down because a coincidence has to match the whole line. [А-Я] - Russian capital letter - gap. \u0020but it's only necessary if you're planning to add a flag to the comment. Pattern.COMMENTS, can be written, though. \, i.e. shielding\d{6} - 6 digits (ay be replaced by [0-9]{6})$ - End of line.Java-code:Boolean matched = s.matches("[А-Я] \\d{6}"); // Если в строке разрешается простой пробел Boolean matched = s.matches("[А-Я]\\s\\d{6}"); // Если в строке разрешены любые пробельные символы
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The main problem is . There's no sign of a new line. In addition, the design used is too cumbersome, it is sufficient to to limit itself to one set of forward vision:/^(?![^]*\s{2})/ ^(?![^]*\s{2}) It's the beginning of the line.^and then, if there are 0 and more symbols followed by 2 of any missing symbols, reverts false♪If you need to cut off the matches to 2+ "vertical" bypassing symbols, use them./^(?![^]*[^\S\r\n]{2})/ [^\S\r\n] only finds "horizontal" underwear symbols (with the exception of some non-typical shifts to a new line, but this scale is most often sufficient).
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You have a little confusion: you're using it. match(), to search()with the flag. gwithout, therefore, different kinds of results fall into an alert, turning the masses into a line. Try to respect uniformity. And if that's allowed, use it. console.log()that'll be more clear.Don't forget to screen special symbols in rows and regular expressions.The strangeness of the result in the first assignment from the fact that the regular will be the same as the zero symbols, from this large number of empty lines to the results.let myRe = /\\*\**/g; let str = "djfhhj\\****jjd"; console.log(str.match(myRe)); myRe = /(.|\n){5}/g; // варианты: /[^]{5}/g или /[\s\S]{5}/g или /.{5}/gs str = "hello world"; console.log(str.match(myRe)); myRe = /(?<=^|\s)(\S+)(?:\s+\1(?=\s|$))+/g; str = "hello my my my friend"; console.log(str.match(myRe));
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If you want to test the whole line http://www.fileformat.info/info/unicode/category/Zs/list.htm and without regular expressions by https://developer.mozilla.org/ru/docs/Web/JavaScript/Reference/Global_Objects/String/Trim :var s = "\u0020\u00A0\u1680\u2000\u2001\u2002\u2003\u2004\u2005\u2006\u2007\u2008\u2009\u200A\u202F\u205F\u3000\t"; console.log("Исходная строка: '", s, "'"); if (s.trim() === '') { console.log('Строка содержит только пробельные символы!'); }If you really need a regular expression, use it. https://ru.stackoverflow.com/questions/734478/%D0%98%D0%B3%D0%BD%D0%BE%D1%80%D0%B8%D1%80%D0%BE%D0%B2%D0%B0%D1%82%D1%8C-s-t-n-javascript-%D1%80%D0%B5%D0%B3%D1%83%D0%BB%D1%8F%D1%80%D0%BD%D1%8B%D0%B5-%D0%B2%D1%8B%D1%80%D0%B0%D0%B6%D0%B5%D0%BD%D0%B8%D1%8F#comment1117510_734478 ♪If it is necessary to check whether the line is composed only of specific missing symbols, such as tabulation, carnet transfer and the new line, a symbolic class may be used:if (/^[\t\r\n]+$/.test(s)) { /* строка содержит только \t, \n и/или \r */ } Here.^ - starting line[\t\r\n]+ - one or more signs of tabulation, relocation of the carriage$ - End of line.Class [\t\r\n] can be replaced by \sif all the missing symbols are to be checked.
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It's not one symbol, it's two underground:string.split('\"')[1] or "to take everything between the barrel and the stalker."string[/(?<=")[^\]+/]
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^starting line[A-Za-z0-9]one Latin letter or number[Grunts]sign, point or less zero or more times[a-zA-Z0-9]+letter or number 1 or more([_-]?[a-zA-Z0-9]+)*total combination 0 or more@dog(sighs) [A-Za-z0-9]+)letter or number 1 or more(grunts) [.- ]? [a-zA-Z0-9]+)*See previous example♪Point(sighs) [A-Za-z] {2,}Latin letter two or more times$US$End of line
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The option of commenting would indeed be useful:^[a-zA-Z]*[A-Z]{2}$
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Expanding iqanansoft's response, you should take the MessageBox out of the loop and replace it with a variable, also investing the condition. Also, I like to use the Equals directly on the string (or the char, as in this case). Something like that:var result = "El número es binario"; foreach (var item in texbox.Text) { if(Not item.Equals('1') AndAlso Not item.Equals('0')) { result = "El número no es binario"; break; } } MessageBox.Show(result); Add an explanation of why your code fails:As you have been told, it is convenient to compare text strings or characters with the Equals methodEven using the Equals method, if you compare a 'item' with the '1' character and also with the '0' character it is impossible for both to be True. By using the & fake (AND) to combine them, that condition will never be metIf for whatever that condition could be given, it would show a messagebox for every invalid character to find, which makes no sense. Hence the break, to finish the execution once an invalid character is found
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I created a regular expression that identifies if a string is a phone, taking into account the following cases:0800 phonesnumber of operators and services like 10315 and 190phones represented with or without parenthesesaccepts operator represented as 0xx11phones with or without the separators [.-] ignores phones started with 0 if you do not have DDD (e.g. 0999-9999 is not accepted, but 0xx11 9123-1234 is)It was a little big and hard to read humanly, but it serves my projects:/^1\d\d(\d\d)?$|^0800 ?\d{3} ?\d{4}$|^($0?([1-9a-zA-Z][0-9a-zA-Z])?[1-9]\d$ ?|0?([1-9a-zA-Z][0-9a-zA-Z])?[1-9]\d[ .-]?)?(9|9[ .-])?[2-9]\d{3}[ .-]?\d{4}$/gm You can see the expression in use http://regexr.com/3dkvr .
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It's not the right way to decide. It's like this:^((^| )(.)(?!.* \3)[^ ]+){2,}$ Verification:"складеною випадковою подією\n\ складена випадкова подія\n\ складеною подією\n\ складена подія\n\ випадковою подією\n\ випадкова подія\n\ випадковою складеною подією\n\ випадкова складена подія\n\ \ складеною складеною подією\n\ складена складена подія\n\ випадковою випадковою подією\n\ випадкова випадкова подія\n\ подією\n\ подія".split("\n").map(function(s) { return /^((^| )(.)(?!.* \3)[^ ]+){2,}$/.test(s); }) // [true, true, true, true, true, true, true, true, false, false, false, false, false, false]

Establish regular expression

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