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  • SubSonic isn't generating MySql foreign key tables

    - by keith
    I two tables within a MySql 5.1.34 database. When using SubSonic to generate the DAL, the foreign-key relationship doesn't get scripted, ie; I have no Parent.ChildCollection object. Looking inside the generated DAL Parent class shows the following; //no foreign key tables defined (0) I have tried SubSonic 2.1 and 2.2, and various MySql 5 versions. I must be doing something wrong procedurally - any help would be greatly appreciated. This has always just worked 'out-the-box' when using MS-SQL. TABLE `parent` ( `ParentId` INT(11) NOT NULL AUTO_INCREMENT, `SomeData` VARCHAR(25) DEFAULT NULL, PRIMARY KEY (`ParentId`) ) ENGINE=INNODB DEFAULT CHARSET=latin1; TABLE `child` ( `ChildId` INT(11) NOT NULL AUTO_INCREMENT, `ParentId` INT(11) NOT NULL, `SomeData` VARCHAR(25) DEFAULT NULL, PRIMARY KEY (`ChildId`), KEY `FK_child` (`ParentId`), CONSTRAINT `FK_child` FOREIGN KEY (`ParentId`) REFERENCES `parent` (`ParentId`) ) ENGINE=INNODB DEFAULT CHARSET=latin1;

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  • Integrating TFS and MySQL

    - by user294043
    We are developing an application with Visual Studio 2008 and TFS. Our database is a MySQL DB. As we develop we keep the new queries that need to be applied to the database of our new release as the New Version Update Queries. Right now I'm keeping them in a simple text file (which is a painful task!). I know that TFS integrates with MSSQL and makes this job very easy. I've already asked our consultant from Microsoft if there is any way to integrate TFS and MySQL, and his answer was "NO". So I was wondering if anyone knows any smart way of handling this issue?

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  • Unix timestamp and mysql date: birthdate

    - by Mikk
    Hi, I have a really basic question concerning unix timestamp and mysql date. I'm trying to build a small website where users can register and fill in their birthdate. Problem is that unix starts with Jan 01 1970. Now if i calculate age for users, form dates like date('m.d.Y', $unix_from_db) and so on it will fail with users older that 40 years, right? So what would be the rigth way for doing this. Sorry, for basic question like this, but I'm inexperienced with php and mysql.

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  • MySQL with Java: Open connection only if possible

    - by emempe
    I'm running a database-heavy Java application on a cluster, using Connector/J 5.1.14. Therefore, I have up to 150 concurrent tasks accessing the same MySQL database. I get the following error: Exception in thread "main" com.mysql.jdbc.exceptions.jdbc4.MySQLNonTransientConnectionException: Too many connections This happens because the server can't handle so many connections. I can't change anything on the database server. So my question is: Can I check if a connection is possible BEFORE I actually connect to the database? Something like this (pseudo code): check database for open connection slots if (slot is free) { Connection cn = DriverManager.getConnection(url, username, password); } else { wait ... } Cheers

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  • How to Enable Full-Text Index on Sql Server 2008 Table

    - by michaeldelorenzo
    Not sure what's happening with this, but here's my question. I have a Sql Server 2008 database that I need to be able to do full-text indexing/searching but when I try to setup my indices on the table, I get the following: I've tried running this stored procedure on my database and it's successful: EXEC sp_fulltext_database @action = 'enable' But I still get the above window and my full-text searches don't return any results when they should. What am I missing?

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  • MySQL insert at end of table

    - by Theopile
    Hello, I am using MySQL and PHP. I am having a problem with inserting a new item at the end of my table. When I insert the new item appears after the last created item, but I want it to be entered at the bottom of the table. Suppose that I have a table id=int,Primary Key and album=string and the table is: Wrath Crack The Skye Enter Shikari What would the MySQL query be if php variable $album=myAlbum was to be inserted next, at the end of the table, and with the appropriate id? Thanks

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  • Question about Radio button/PHP/MySQL

    - by Marcelo
    Hi, I'm an engineering student and I'm developing a simple software based on HTML, PHP and mysql. I learned this topics on w3schools. I know only the basics. I tried to search about this in this website but I found questions about PHP, MySQL and HTML radio buttons which were much more complex than I need and that I could understand. Sorry for the English. (Q1) Ex: $email=$_REQUEST['email'] , in this case the input is text, if it where like a radio button for ex: sex: male or female, how would it be? (Q2) what would be the type of this field (for exemple sex in question 1) in the database: text, int, varchar ? Thanks for the attention

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  • MySQL different versions other results.

    - by kuba
    hey, i have 2 version of mysql on windows 5.1.39-community and on linux 5.1.39-log i execute a query: SELECT `o`.`idOffer`, `o`.`offer_date`, `p`.`factory`, `c`.`short` AS `company`, `s`.`name` AS `subcategory`, `ct`.`name` AS `category`, count( (select count(1) from product where idProduct=idOffer group by idOffer) ) as b FROM `Offer` AS `o` LEFT JOIN `Product` AS `p` ON o.idOffer = p.idOffer LEFT JOIN `company` AS `c` ON o.company = c.id LEFT JOIN `Subcategory` AS `s` ON s.idSubcategory = o.idSubcategory LEFT JOIN `Category` AS `ct` ON ct.idCategory = s.idCategory WHERE (o.idOffer = p.idOffer) GROUP BY `o`.`idOffer` on windows it works as it suppose, but on linux it says: ERROR 1242 (21000): Subquery returns more than 1 row is it any way to get it worked on linux without any mysql updates/downgrades ?

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  • Finding shared list IDs in a MySQL table using bitwise operands

    - by landons
    I want to find items in common from the "following_list" column in a table of users: +----+--------------------+-------------------------------------+ | id | name | following_list | +----+--------------------+-------------------------------------+ | 9 | User 1 | 26,6,12,10,21,24,19,16 | | 10 | User 2 | 21,24 | | 12 | User 3 | 9,20,21,26,30 | | 16 | User 4 | 6,52,9,10 | | 19 | User 5 | 9,10,6,24 | | 21 | User 6 | 9,10,6,12 | | 24 | User 7 | 9,10,6 | | 46 | User 8 | 45 | | 52 | User 9 | 10,12,16,21,19,20,18,17,23,25,24,22 | +----+--------------------+-------------------------------------+ I was hoping to be able to sort by the number of matches for a given user id. For example, I want to match all users except #9 against #9 to see which of the IDs in the "following_list" column they have in common. I found a way of doing this through the "SET" datatype and some bit trickery: http://dev.mysql.com/tech-resources/articles/mysql-set-datatype.html#bits However, I need to do this on an arbitrary list of IDs. I was hoping this could be done entirely through the database, but this is a little out of my league. Any bit gurus out there? Thanks, Landon

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  • Printer redirection on server 2003

    - by user137841
    On windows server 2003 when one user connects to the server via RDP the default printer of the server for her profile does not change to the redirected printer of the session. This only happens with the one user all the other users default printers defaults to their session printer automatically. I tried the following solution but there was no \Terminal Server\Printer Redirection in gpedit.msc http://technet.microsoft.com/en-us/library/cc731963(v=ws.10).aspx Computer Configuration\Policies\Administrative Templates\Windows Components\Terminal Services\Terminal Server\Printer Redirection Is there a different place to check the Printer Redirection?

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  • How to SET tiggers 0 in MySQL?

    - by Grijesh Chauhan
    In my MySQL database I have some Triggers ON DELETE and ON INSERT. Some time I need to switched-off my some Triggers, And I have to DROP e.g. DROP TRIGGER IF EXISTS hostgroup_before_insert // and reinstall. Is there any shortcut to SET triggers hostgroup_before_insert = 0 like we have for foreign keys mysql> SELECT version(); +-------------------------+ | version() | +-------------------------+ | 5.1.61-0ubuntu0.10.10.1 | +-------------------------+ 1 row in set (0.00 sec) I am new learner and I could not find regarding this on web.

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