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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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  • Organize array in PHP from mysql

    - by Matthew Carter
    Hi i have a social networking website. what i want it to do is pull out my friends status updates. basically what it does is i have a mysql query that pulls out all of my friends and in that while loop there is another mysql query that pulls out the status's from my friends. i want it to be in order of date but since its one while loop in another what it does is pull out all status's from friend 1 then 2 then 3 and not in order by date. i even tried ORDER BY DATE but that just ordered it by date within the friend.. my thought is that i could putt it all in an array and friends is one thing and the values is the stats. then just sort by values would this work and how could i do it. THANKS SO MUCH

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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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  • 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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  • Mysql query taking too much time

    - by aditya
    I have problem related to mysql database. i am linux webserver admin and i am facing a problem with a mysql query. The database is very small. I tried to track in logs and found that a query is taking minimum 5 sec to respond . The first page of site is coming from the database. Client are using cms. when the server gets some number of hits database server starts to give response very slowly and wait time increases from 5 sec to several seconds. I checked slow query logs { Query_time: 11.480138 Lock_time: 0.003837 Rows_sent: 921 Rows_examined: 3333 SET timestamp=1346656767; SELECT `Tender`.`id`, `Tender`.`department_id`, `Tender`.`title_english`, `Tender`.`content_english`, `Tender`.`title_hindi`, `Tender`.`content_hindi`, `Tender`.`file_name`, `Tender`.`start_publish`, `Tender`.`end_publish`, `Tender`.`publish`, `Tender`.`status`, `Tender`.`createdBy`, `Tender`.`created`, `Tender`.`modifyBy`, `Tender`.`modified` FROM `mcms_tenders` AS `Tender` WHERE `Tender`.`department_id` IN ( 31, 33, 32, 30 ); } Every line in the log is same only there is diff in Query time. Is there any way tweak the performance?

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  • Create database in Shell Script - convert from PHP

    - by snaken
    I have the following PHP code that i use to create a databaase and grant permissions to a user: $con = mysql_connect("IP.ADDRESS","user","pass"); mysql_query("CREATE DATABASE ".$dbuser."",$con)or die(mysql_error()); mysql_query("grant all on ".$dbuser.".* to ".$dbname." identified by '".$dbpass."'",$con) or die(mysql_error()); I want to perform these same actions but from within a shell script. Is it just something like this: MyUSER="user" MyPASS="pass" MYSQL -u $MyUSER -h -p$MyPASS -Bse "CREATE DATABASE $dbuser;' MYSQL -u $MyUSER -h -p$MyPASS -Bse "GRANT ALL ON ${DBUSER}.* to $DBNAME identified by $DBPASS;"

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  • Splitting a MySQL DB in two may ease server from "Too many connetions"? I don't think so

    - by Petruza
    I was requested to split a MySQL in two, it's kind of a horizontal partition, in which some rows correspond to one site, and some other correspond to another site. But they want to split it in two DBs in the same MySQL server. I'm no DB expert but I guess keeping them in the same MySQL server with the same amount of memory and processor and the same platform won't improve things. What we're trying to avoid is the "Too many connections" problem.

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  • MYSQL Event Scheduler DELIMITER using PHP

    - by user1440918
    I'm having an issue with my PHP code trying to create events within MySQL. I begin with creating a string like this: $sql="DELIMITER $$ CREATE EVENT `$test_name` ON SCHEDULE EVERY $time1 $sched2 STARTS '$start_date $start_time' DO BEGIN "; $sql .="INSERT INTO blah (foo,bar); "; $sql .="END$$ DELIMITER ;" mysql_query($sql,$dbh); But I keep getting Syntax Errors starting with DELIMITER $$ CREATE EVENT. Without the semicolon behind (foo,bar); the event triggers with a unexecuted payload. Any ideas on where I'm going wrong? Thanks!

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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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  • 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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  • Table Design in mysql

    - by RIDDHI
    Hi everyone, I need to create one table, Description : i need to create table based on schedule like daily, weekly & monthly, columns are like : sno, startdate, enddate, day, scheduletype For example i ll take weekly data, for my point of view : From sunday to saturday (1 - 7 )Id i create.... So lots of posibilities are creates like (1,2)(1,3) ..(1,2,3)....up to n....this twise posibility only but that will created up to 7 posibility in one. so how can i store this posibility in mysql database? If any one have an issue get back to me... Thanks in advanced!!!! Riddhi

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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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  • mysql replace matching but not changing

    - by alex
    I've used mysql's update replace function before, but even though I think I'm following the same syntax, I can't get this to work-it matches the rows, but doesn't replace. Here's what I'm trying to do: mysql> update contained_widgets set preference_values = replace(preference_values, '<li><a_href="/enewsletter"><span class="not-tc">eNewsletter</span></a></li>', '<li><a_href="/enewsletter"><span class="not-tc">eNewsletter</span></a></li> <li> <a_href="/projects"><span class="not-tc">Projects</span></a></li>'); Query OK, 0 rows affected (0.00 sec) Rows matched: 77 Changed: 0 Warnings: 0 I don't see what I'm missing. Any help is appreciated. I edited "a " to "a_" because the site thinks I'm posting spam links otherwise.

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  • MySQL Removing Some Foreign keys

    - by Drew
    I have a table whose primary key is used in several other tables and has several foreign keys to other tables. CREATE TABLE location ( locationID INT NOT NULL AUTO_INCREMENT PRIMARY KEY ... ) ENGINE = InnoDB; CREATE TABLE assignment ( assignmentID INT NOT NULL AUTO_INCREMENT PRIMARY KEY, locationID INT NOT NULL, FOREIGN KEY locationIDX (locationID) REFERENCES location (locationID) ... ) ENGINE = InnoDB; CREATE TABLE assignmentStuff ( ... assignmentID INT NOT NULL, FOREIGN KEY assignmentIDX (assignmentID) REFERENCES assignment (assignmentID) ) ENGINE = InnoDB; The problem is that when I'm trying to drop one of the foreign key columns (ie locationIDX) it gives me an "ERROR 1025 (HY000): Error on rename" error. How can I drop the column in the assignment table above without getting this error?

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  • Insert record into mysql db with Entity Framework

    - by sanfra1983
    Hi, the problem is that it will insert a new record in a mysql table, I have already done the mapping of the mysql db and I have already done tests returning data and everything works. Now I read from a file, where there are queries written, I have them run me back and the result of true or false based on the final outcome of single query written to the file. Txt; I did this: using (var w = new demotestEntities ()) ( foreach (var l listaqueri) ( var p = we.CreateQuery <category> (l); we.SaveChanges (); result = true; ) ) but it does not work, I sense that it returns no errors, but neither the result given written in the query. txt file is as follows: INSERT INTO category (id, name) VALUES (null, 'test2') anyone can help me?

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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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  • Selecting two field in MySQL with PHP

    - by Crays
    Hi guys, i'm relatively new to php and mysql and would like to know how to select two value in mysql with php. What i have is $query = sprintf("SELECT COUNT(id) FROM table WHERE UPPER(username) = UPPER('%s') AND password='%s'"... in this case, i'm only selecting and count if the id exist and i use list($count) = mysql_fetch_row($result); if($count == 1) and by using cookies, i would like to retrieve two value from the database, namely user (the user's name) and power (which has value of 1,2 or 3, indicating the menu they would be able to see) basically it is to differentiate if you're admin or normal user, but i wonder if i could do SELECT COUNT(id) AND power FROM table WHERE ... is this possible? or is there any other way? Please guide me, thanks.

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