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Nomenclatures in SQL in 1C

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nishika last edited by

How do 1c trade keep sql images from the nomenclature and how do we get them from there?
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Analeea last edited by

The pictures in 1C are stored in a special storage container. This Container is showing them in some form. And then there's a SQL record. Like a field in SQL, with a image.
Here's some example of getting a picture from the SQL field. The truth is an example on C#. But we can figure it out. http://igorus1977.ya.ru/replies.xml?item_no=1462
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What do I do if I do not find the Sql Server Configuration Manager on Windows 10 for SQL Server Management Studio v17.0 RC1?
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The SQL Server configuration manager is is usually accessible as a complement to the administration console (microsoft management console: mmc.exe).You can run it directly from the dialog box execute (which you can in turn invoke with the combination of keys Windowsr.In that dialogue, in the edit Open writes:SQLServerManager15.msc for SQL Server 2019SQLServerManager14.msc for SQL Server 2017SQLServerManager13.msc for SQL Server 2016SQLServerManager12.msc for SQL Server 2014SQLServerManager11.msc for SQL Server 2012SQLServerManager10.msc for SQL Server 2008svrnetcn.exe for SQL Server 2000Then press Enter.More information in the MSDN: http://technet.microsoft.com/en-us/library/ms174212.aspx
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How do I get the highest number with having?
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Oscar,With this consultation you will get the result you are looking for:SELECT fecha, cuenta, MAX(estatus) estatus FROM d WHERE estatus IN (1,4,8) GROUP BY fecha, cuenta ORDER BY 1, 2; FECHA CUENTA ESTATUS 2020-05-02 345 1 2020-05-03 345 8 2020-05-03 678 4 In a query first the condition of the WHERE is evaluated, therefore the universe that is observed when grouping is formed only on those records with status (1, 4, 8). Records that are in another state are excluded at the beginning.In this other query (you don't get what you're looking for.), not having a filter in the WHERE, participates all records. But only those whose maximum status values are within (1, 4, are part of the result. SELECT fecha, cuenta, MAX(estatus) estatus FROM d GROUP BY fecha, cuenta HAVING MAX(estatus) IN (1,4,8) ORDER BY 1, 2; FECHA CUENTA ESTATUS 2020-05-03 345 8 2020-05-03 678 4 This is not very orthodox but it will help you think of the HAVING section as a WHERE of the GROUP BY.I hope it serves you. Greetings.
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Is there a T-SQL equivalent in the PL/SQL of the AC-oriented content?
Software Programming • sql oracle plsql translated • • Burma

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There you go.select listagg(dname, ',') within group (order by dname) result from dept; RESULT ACCOUNTING,OPERATIONS,RESEARCH,SALES Here. https://dbfiddle.uk/?rdbms=oracle_18&fiddle=a5a56e1900e39f52f07f5886b77a37ee ♪Free translation https://stackoverflow.com/a/63712376 from the participant https://stackoverflow.com/users/9097906/littlefoot
SQL
Software Programming • c sharp linq lin • • Alberto

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FirstSame as ID and dateThese items must be grouped.Determining whether each group meets the condition in this caseWHERE コード IN ('1234', '1235')and limit the number of groupsCOUNT(DISTINCT コード)I think that you should have two types of code.So SQL to get the group key is as follows.SELECT ID, 日付, 連番 FROM tableA WHERE コード IN ('1234', '1235') GROUP BY ID, 日付, 連番 HAVING COUNT(DISTINCT コード) = 2 Use this command-line statement for sub-inquiriesJOINHomeINHomeEXISTSYou can realize the desired SQL by with it.ExampleINNER JOINWhen usingSELECT * FROM tableA t INNER JOIN ( SELECT ID, 日付, 連番 FROM tableA WHERE コード IN ('1234', '1235') GROUP BY ID, 日付, 連番 HAVING COUNT(DISTINCT コード) = 2 ) k ON t.ID = k.ID AND t.日付 = k.日付 AND t.連番 = k.連番 WHERE コード IN ('1234', '1235') likeLINQ implementation is also possible with the same policy.
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SQL LEFT JOIN
Software Programming • mysql sql join • • fbio

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The solution is quite quickly:SELECT task.id AS id, task.title AS title, complete.use_tip AS free_tip, complete.use_paid_tip AS paid_tip, (complete.task_id IS NOT NULL) AS complete FROM task LEFT JOIN complete ON (task.id=complete.task_id AND complete.user_id=3)
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Relationship (Database) (Sql)
Software Programming • sql databank • • jules

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You should always take the strongest relationship, in case the N. Don't forget that the 'lotacao' rhombus serves only to guide your head and realize what kind of relationship exists between the two tables. The ratio of 1 to N prevails because the N always weighs more.If you had (N:1) -- (1:N) would be (N:N) and would have to create an associative table because it would be a relation of many to many. If he had (1:N) -- (1:1) he would (N:1).I hope I helped!
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How to make Postgres create a file to populate a bank?
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To generate commands insert of the data that contains in the bank you can do as follows:Select Plain format In the Dump Options 1 tab, mark "Data" In the Dump Options 2 tab, mark "User Insert Commands" I suggest using notepad++ to open
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Guidance with sql
Software Programming • mysql sql base de datos • • Trenton

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Why don't you make a UNION? https://dev.mysql.com/doc/refman/8.0/en/union.html SELECT artista, nombre_disco, año FROM CDs WHERE NOT artista = "Varios" ORDER BY nombre_disco, año UNION SELECT artista, nombre_disco, año FROM CDs WHERE artista = "Varios" ORDER BY artista, año Watch the NOT in the first WHERE.
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I can't get a sql.
Software Programming • sql • • Pearlaqua

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select mm,cnt as curr,prev,round(100-cnt*100.00/prev,2) prc from ( select mm,cnt,lag(cnt) over (order by mm) prev from ( select year(creationDate)*100+month(creationDate) as mm, count(1) as cnt from posts where PostTypeId between 1 and 2 group by year(creationDate)*100+month(creationDate) ) A ) A where cnt < prev order by mm
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What is the best way to make a current contract selection and previous employee contract in the same row?
Software Programming • sql postgresql oracle orac languagele 11g • • obi

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Functions type windowingThere are windowing type functions to do these tasks: https://www.postgresql.org/docs/current/static/functions-window.html Function lag Gives you the top line and lead The bottom line. If you just want to show the last line (with the top one) you have to put a select within another to filter at the end. The example goes to understand:select * from ( select empleados.*, ciudades.ciudad, contratos.contract_id, contratos.start_date, contratos.end_date, lag(contratos.contract_id) over (partition by empleados.employee_id order by contratos.start_date) as contrato_anterior, lag(contratos.start_date) over (partition by empleados.employee_id order by contratos.start_date) as contrato_anterior_start_date, lag(contratos.end_date) over (partition by empleados.employee_id order by contratos.start_date) as contrato_anterior_end_date from contratos inner join empleados on contratos.employee_id=empleados.employee_id left join ciudades on ciudades.city_id = empleados.city_id ) select_base_con_todo_junto where end_date is null; full versionI put a full version with the inserts and create table in: http://rextester.com/EPM24088
MySql 1052 Column 'id' in IN/ALL/ANY subquery is ambiguous
Software Programming • php mysql sql yii2 • • briley

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On request, the id field is at least 2x tables (item and currency)♪ WHERE You need to point out the id from which table you want to filter.If this is the item, the request shall be: SELECT item.* FROM item JOIN price ON price.item = item.id JOIN currency ON currency.id = price.currency WHERE item.`id` IN (SELECT `item_id` FROM `item_code` WHERE `number` IN (SELECT `id` FROM `number` WHERE `mark`='3')) ORDER BY price.price * currency.course LIMIT 20
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You're probably getting the names of the because of this line:String[] headers = new String[] {DatabaseHelper.SITE_NAME_COLUMN, DatabaseHelper.LOGIN_COLUMN,DatabaseHelper.PASSWORD_COLUMN}; You create a set of lines where you place the name of the columns, and then remove them. A data should be obtained from the request (using cursor).Example:cursor.moveToFirst(); String item_content = cursor.getString(cursor.getColumnIndex(DatabaseHelper.LOGIN_COLUMN)); login.setText(item_content);
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Create sql database connection in a 8 MySQL and SQL Server drupal module
Software Programming • mysql drupal drupal8 • • rielyn

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In the end the most comfortable way, and that allows me to do what I want, has been the following code:<?php /** * @file * Contains \Drupal\edicion_slider\Controller\edicionslider. */ namespace Drupal\edicion_slider\Controller; class edicionslider { public function content() { include "llamadabdatossli.php"; return array( '#type' => 'markup', '#markup' => basedatosslider(), '#attached' => [ 'library' => [ 'edicion_slider/slider_admin', ], ], ); } } ?> I mounted the module normally, but in the php file that goes inside the controller what I did was call a function located in another file, in this case calledbdatossli.php. The Drupal 8 system is not bad, but there are things I prefer to manage in a more direct way, but without having to touch the chore. Therefore the call to the database, image insertion, editing and display is done all through a file, the one that mentions before, which, through PDO and prepare statements, manages the database and inputs that the user can make. Of course, if Drupal 8 allows you to do what you want to do, the module should be done as indicated in the tutorials, but in this case, that wanted more precision, more blocks and that a single block can show more content on condition, it has been the best way for me.
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Translate SQL query to Eloquent
Software Programming • php mysql sql eloquent • • Anderson

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First of all I see, there is a syntax error in DB::table('vips_products AS Product), it is missing to close that string. Second, you're using Query Builder but not Eloquent. If what you need is to get that query in Query Builder would be like this:$query = DB::table('vips_categories') ->join('vips_products','vips_categories.parent_id', '=', 'vips_products.id') ->join('vip_archivos', 'vip_archivos.id', '=', 'vips_products.img') ->where('vips_categories.parent_id', '1') ->select(['*']);
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Get the sum of the authors' books with Mysql
Software Programming • mysql sql • • Nykeriab

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Use the query below:select u.nome, sum(l.preco) from usuario u, livro l, carrinho_de_compras as c where c.id_user = u.id_user and c.id_livro = l.id_livro group by u.nome; Removed inner join and put validation on where; mysql was "losing" with validation in the clause on.

Nomenclatures in SQL in 1C

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