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  • mysql real escape string error

    - by user547995
    Code mysql_query("INSERT INTO Account(User, Pw, email) VALUES('mysql_real_escape_string($_POST[user])', '$pw','mysql_real_escape_string($_POST[email]) ) ") or die(mysql_error()); Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''mysql_real_escape_string(123) )' at line 1 Please help

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  • Require help in Writing Query

    - by harigm
    The following image have been uploaded to show what I am trying to do and what I wanted out of it Can any one help me write the Query to get the results what I want Please check the following SELECT * FROM KPT WHERE PROPERTY_ID IN (SELECT PROPERTY_ID FROM khata_header WHERE DIV_ID = 3 and RECORD_STATUS = 0) and CHALLAN_NO > 42646 The above is the query I have written and I have got the following result set ID CHALLAN_NO PROPERTY_ID SITE_NO TOTAL_AMOUNT ----- ------------- -------------- ------------------- --------------- 1242 42757 3103010141 296 595 1243 63743 3204190257 483 594 1244 63743 3204190257 483 594 1334 43395 3217010223 1088 576 1421 524210 3320050416 (null) (null) 1422 524210 3320050416 (null) (null) 1560 564355 3320021408 (null) (null) 1870 516292 3320040420 (null) (null) 1940 68357 3217100104 139 1153 1941 68357 3217100104 139 1153 2002 56256 3320100733 511 4430 2003 56256 3320100733 511 4430 2004 66488 3217040869 293 3094 2005 66488 3217040869 293 3094 2016 64571 3217040374 (null) (null) 2036 523122 3320020352 (null) (null) 2039 65682 3217040021 273 919 In my resultset, I am getting the PropertyId repeated, since there are multilple entries, How Can I know How many have been repeated What are those Property Id which have repeated more than 2 times. Little Back ground about the tables are PROPERTY_ID is the FK in the KPT PROPERTY_ID is the PK in KH I am writing a subquery to get the Result, so I am stuck I dont know how to get my results Please help

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  • Select random row from MySQL (with probability)

    - by James Simpson
    I have a MySQL table that has a row called cur_odds which is a percent number with the percent probability that that row will get selected. How do I make a query that will actually select the rows in approximately that frequency when you run through 100 queries for example? I tried the following, but a row that has a probability of 0.35 ends up getting selected around 60-70% of the time. SELECT * FROM table ORDER BY RAND()*cur_odds DESC

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  • MySQL MyISAM table performance... painfully, painfully slow

    - by Salman A
    I've got a table structure that can be summarized as follows: pagegroup * pagegroupid * name has 3600 rows page * pageid * pagegroupid * data references pagegroup; has 10000 rows; can have anything between 1-700 rows per pagegroup; the data column is of type mediumtext and the column contains 100k - 200kbytes data per row userdata * userdataid * pageid * column1 * column2 * column9 references page; has about 300,000 rows; can have about 1-50 rows per page The above structure is pretty straight forwad, the problem is that that a join from userdata to page group is terribly, terribly slow even though I have indexed all columns that should be indexed. The time needed to run a query for such a join (userdata inner_join page inner_join pagegroup) exceeds 3 minutes. This is terribly slow considering the fact that I am not selecting the data column at all. Example of the query that takes too long: SELECT userdata.column1, pagegroup.name FROM userdata INNER JOIN page USING( pageid ) INNER JOIN pagegroup USING( pagegroupid ) Please help by explaining why does it take so long and what can i do to make it faster. Edit #1 Explain returns following gibberish: id select_type table type possible_keys key key_len ref rows Extra 1 SIMPLE userdata ALL pageid 372420 1 SIMPLE page eq_ref PRIMARY,pagegroupid PRIMARY 4 topsecret.userdata.pageid 1 1 SIMPLE pagegroup eq_ref PRIMARY PRIMARY 4 topsecret.page.pagegroupid 1 Edit #2 SELECT u.field2, p.pageid FROM userdata u INNER JOIN page p ON u.pageid = p.pageid; /* 0.07 sec execution, 6.05 sec fecth */ id select_type table type possible_keys key key_len ref rows Extra 1 SIMPLE u ALL pageid 372420 1 SIMPLE p eq_ref PRIMARY PRIMARY 4 topsecret.u.pageid 1 Using index SELECT p.pageid, g.pagegroupid FROM page p INNER JOIN pagegroup g ON p.pagegroupid = g.pagegroupid; /* 9.37 sec execution, 60.0 sec fetch */ id select_type table type possible_keys key key_len ref rows Extra 1 SIMPLE g index PRIMARY PRIMARY 4 3646 Using index 1 SIMPLE p ref pagegroupid pagegroupid 5 topsecret.g.pagegroupid 3 Using where Moral of the story Keep medium/long text columns in a separate table if you run into performance problems such as this one.

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  • SQL query using information from 4 tables (not all directly linked)

    - by Yvonne
    I'm developing a simple classroom system, where teachers manage classes and their subjects. I have 2 levels of access in my teachers table, assigned by an integer (1 = admin, 2 = user)... Meaning that the headteacher is the admin :) A teacher (of level 1) can have have many classes and a class can have many teachers (so I have 'TeachersClasses' table). A class can have many subjects, and a teacher can have many subjects. Basically, I'm attempting a query to display the admin teacher's (level 1) subjects. However, only teachers with a level of 2, are directly related to a subject, which is set by the admin user. The headteacher can view all of their subjects via the classroom, but I cannot get all of the subjects to be displayed on one page, instead I can only get the subjects to appear under a specific classroom right now... This is what I have so far, which is returning nothing. (I'm guessing this may require an SQL clause more advanced that 'INNER JOIN' which is the only join type I am familiar with, and thought it would be enough! $query = "SELECT subjects.subjectid, subjects.subjectname, subjects.subjectdetails, classroom.classid, classroom.classname FROM subjects INNER JOIN classroom ON subjects.subjectid = classroom.classid INNER JOIN teacherclasses ON classroom.classid = teacherclasses.classid INNER JOIN teachers ON teacherclasses.teacherid = teachers.teacherid WHERE teachers.teacherid = '".intval( $_SESSION['SESS_TEACHERID'] )."'"; In order for all subjects related to the headteachers class to be displayed, I'm gathering that all of my tables will need to be called up here? Thanks for any help! Example output: subject name: maths // teacher: mr smith // classroom: DG99 x10 for all the subjects associated with the headteachers classrooms :)

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  • Create a Cumulative Sum Column in MySQL

    - by Kirk
    I have a table that looks like this: id count 1 100 2 50 3 10 I want to add a new column called cumulative_sum, so the table would look like this: id count cumulative_sum 1 100 100 2 50 150 3 10 160 Is there a MySQL update statement that can do this easily? What's the best way to accomplish this?

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  • Advice on how to complete specific MySQL JOIN

    - by Tim
    Hello, I have a mysql table jobs. This is the basic structure of jobs. id booked_user_id assigned_user_id I then also have another table, meta. Meta has the structure: id user_id first_name last_name How can I join these tables so that both booked_user_id and assigned_user_id can access meta.first_name? Thanks for your advice Tim

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  • mysql query for getting all messages that belong to user's contacts

    - by aharon
    So I have a database that is setup sort of like this (simplified, and in terms of the tables, all are InnoDBs): Users: contains based user authentication information (uid, username, encrypted password, et cetera) Contacts: contains two rows per relationship that exists between users as (uid1, uid2), (uid2, uid1) to allow for a good 1:1 relationship (must be mutual) between users Messages: has messages that consist of a blob, owner-id, message-id (auto_increment) So my question is, what's the best MySQL query to get all messages that belong to all the contacts of a specific user? Is there an efficient way to do this?

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  • Saving commands for later re-use in MySQL?

    - by Zombies
    What would be the equivalant in MySQL for: Saving a command for later reuse. eg: alias command1='select count(*) from sometable;' Where then I simply type command 1 to get the count for SomeTable. Saving just a string, or rather part of a command. eg: select * from sometable where $complex_where_logic$ order by attr1 desc; WHere $complex_where_logic$ is something I wish to save and not have to keep writing out

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  • MySQL Workbench - How to synchronize the EER Diagram

    - by Tiago Alves
    I am creating a visual representation of my existing database with MySQL Workbench and I am able to synchronize the models with the "Database - Synchronize Model..." menu. However, every time I synchronize (update) my model, I have to recreate the EER Diagram and rearrange all the tables. Is there a way to update or synchronize the EER Diagram as well? Thanks.

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  • regenerate the ID row in a MySQL Table with a Mysql's Script or PHP

    - by DomingoSL
    Hello, i have a database fill with information of the users who use my webpage. The table as many MySql tables have the ID parameters who is autoincrement. The issue is that when somebody eliminate his account from the site, in the database remain a jump in the sequence that i dont want cuz i have a script who fail if find some jump in the ID. Ex. ID Name PASS 1 Jhon 1234 2 Max 2233 3 Jorge 2232 If Max get out and a new user go in, this is what will happend. ID Name PASS 1 Jhon 1234 4 NewU 1133 3 Jorge 2232 So what is the best way to erase some body from the data base in order to avoid this isuue, or if is not a way, its posible to do a PHP or MySql script who eliminate all the contents in the ID row and regenerate it in order? Thanks A lot! sorry for my english

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  • SQL Query Returning Duplicate Results

    - by Jesse Bunch
    Hi, I've been working out this query now for a while and I thought I had it where I wanted it, but apparently not. There are two records in the database (orders). The query should return two different rows, but instead returns two rows that have exactly the same values. I think it may be something to do with the GROUP BY or derived tables I'm using but my eyes are tired and not seeing the problem. Can any of you help? Thanks in advance. SELECT orders.billerID, orders.invoiceDate, orders.txnID, orders.bName, orders.bStreet1, orders.bStreet2, orders.bCity, orders.bState, orders.bZip, orders.bCountry, orders.sName, orders.sStreet1, orders.sStreet2, orders.sCity, orders.sState, orders.sZip, orders.sCountry, orders.paymentType, orders.invoiceNotes, orders.pFee, orders.shipping, orders.tax, orders.reasonCode, orders.txnType, orders.customerID, customers.firstName AS firstName, customers.lastName AS lastName, customers.businessName AS businessName, orderStatus.statusName AS orderStatus, IFNULL(orderItems.itemTotal, 0.00) + orders.shipping + orders.tax AS orderTotal, IFNULL(orderItems.itemTotal, 0.00) + orders.shipping + orders.tax - IFNULL(payments.totalPayments, 0.00) AS orderBalance FROM orders LEFT JOIN customers ON orders.customerID = customers.id LEFT JOIN orderStatus ON orders.orderStatus = orderStatus.id LEFT JOIN ( SELECT orderItems.orderID, SUM(orderItems.itemPrice * orderItems.itemQuantity) as itemTotal FROM orderItems GROUP BY orderItems.orderID ) orderItems ON orderItems.orderID = orders.id LEFT JOIN ( SELECT payments.orderID, SUM(payments.amount) as totalPayments FROM payments GROUP BY payments.orderID ) payments ON payments.orderID = orders.id

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  • Query MSQL for winners, starting at xth place using SELECT

    - by incrediman
    In my MySQL table Winners, I have a list of people who have won. What I'd like to do is select a list of the names of 10 winners. So what I have right now is this: SELECT name FROM Winners ORDER BY points DESC LIMIT 10 This returns the first 10 winners which is great. But how can I make it (for example) return 10 winners, but starting at 20th place?

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  • SQL Server query replace for MySQL Instruction

    - by pojomx
    Hi, I'm new to SQL Server, and used mysql for a while now... SELECT A.acol, IF(A.acol<0,"Neg","Pos") as Column2 From Table I want to do something like that on SQL Server, but there doesn't exist the IF instruction. How do I replace that if, in a SQL Server 2008 Query?

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  • multi word query in mysql

    - by salmane
    Hi there , in order to make things easier for users i want to add multiple keyword search to my site. so that in the input the user would do something like : " keyword1 keyword 2" ( similar to google for example. would i need to write a code that would parse that string and do queries based on that or is there something built in mysql that could do it?

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  • MySQL foreign key creation with alter table command

    - by user313338
    I created some tables using MySQL Workbench, and then did forward ‘forward engineer’ to create scripts to create these tables. BUT, the scripts lead me to a number of problems. One of which involves the foreign keys. So I tried creating separate foreign key additions using alter table and I am still getting problems. The code is below (the set statements, drop/create statements I left in … though I don’t think they should matter for this): SET @OLD_UNIQUE_CHECKS=@@UNIQUE_CHECKS, UNIQUE_CHECKS=0; SET @OLD_FOREIGN_KEY_CHECKS=@@FOREIGN_KEY_CHECKS, FOREIGN_KEY_CHECKS=0; SET @OLD_SQL_MODE=@@SQL_MODE, SQL_MODE='TRADITIONAL'; DROP SCHEMA IF EXISTS `mydb` ; CREATE SCHEMA IF NOT EXISTS `mydb` DEFAULT CHARACTER SET latin1 COLLATE latin1_swedish_ci ; -- ----------------------------------------------------- -- Table `mydb`.`User` -- ----------------------------------------------------- DROP TABLE IF EXISTS `mydb`.`User` ; CREATE TABLE IF NOT EXISTS `mydb`.`User` ( `UserName` VARCHAR(35) NOT NULL , `Num_Accts` INT NOT NULL , `Password` VARCHAR(45) NULL , `Email` VARCHAR(45) NULL , `User_ID` INT NOT NULL AUTO_INCREMENT , PRIMARY KEY (`User_ID`) ) ENGINE = InnoDB; -- ----------------------------------------------------- -- Table `mydb`.`User_Space` -- ----------------------------------------------------- DROP TABLE IF EXISTS `mydb`.`User_Space` ; CREATE TABLE IF NOT EXISTS `mydb`.`User_Space` ( `User_UserName` VARCHAR(35) NOT NULL , `User_Space_ID` INT NOT NULL AUTO_INCREMENT , PRIMARY KEY (`User_Space_ID`), FOREIGN KEY (`User_UserName`) REFERENCES `mydb`.`User` (`UserName`) ON UPDATE CASCADE ON DELETE CASCADE) ENGINE = InnoDB; SET SQL_MODE=@OLD_SQL_MODE; SET FOREIGN_KEY_CHECKS=@OLD_FOREIGN_KEY_CHECKS; SET UNIQUE_CHECKS=@OLD_UNIQUE_CHECKS; The error this produces is: Error Code: 1005 Can't create table 'mydb.user_space' (errno: 150) Anybody know what the heck I’m doing wrong?? And anybody else have problems with the script generation done by mysql workbench? It’s a nice tool, but annoying that it pumps out scripts that don’t work for me. [As an fyi here’s the script it auto-generates: SET @OLD_UNIQUE_CHECKS=@@UNIQUE_CHECKS, UNIQUE_CHECKS=0; SET @OLD_FOREIGN_KEY_CHECKS=@@FOREIGN_KEY_CHECKS, FOREIGN_KEY_CHECKS=0; SET @OLD_SQL_MODE=@@SQL_MODE, SQL_MODE='TRADITIONAL'; DROP SCHEMA IF EXISTS `mydb` ; CREATE SCHEMA IF NOT EXISTS `mydb` DEFAULT CHARACTER SET latin1 COLLATE latin1_swedish_ci ; -- ----------------------------------------------------- -- Table `mydb`.`User` -- ----------------------------------------------------- DROP TABLE IF EXISTS `mydb`.`User` ; CREATE TABLE IF NOT EXISTS `mydb`.`User` ( `UserName` VARCHAR(35) NOT NULL , `Num_Accts` INT NOT NULL , `Password` VARCHAR(45) NULL , `Email` VARCHAR(45) NULL , `User_ID` INT NOT NULL AUTO_INCREMENT , PRIMARY KEY (`User_ID`) ) ENGINE = InnoDB; -- ----------------------------------------------------- -- Table `mydb`.`User_Space` -- ----------------------------------------------------- DROP TABLE IF EXISTS `mydb`.`User_Space` ; CREATE TABLE IF NOT EXISTS `mydb`.`User_Space` ( `User_Space_ID` INT NOT NULL AUTO_INCREMENT , PRIMARY KEY (`User_Space_ID`) , INDEX `User_ID` () , CONSTRAINT `User_ID` FOREIGN KEY () REFERENCES `mydb`.`User` () ON DELETE NO ACTION ON UPDATE NO ACTION) ENGINE = InnoDB; SET SQL_MODE=@OLD_SQL_MODE; SET FOREIGN_KEY_CHECKS=@OLD_FOREIGN_KEY_CHECKS; SET UNIQUE_CHECKS=@OLD_UNIQUE_CHECKS; ** Thanks!]

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  • How to create sql insert query dynamically in mysql

    - by nectar
    I am creating an application where I am generating pins dynamically based on user's input and storing them into mySql database. $sql = "INSERT INTO tblpin ('pinId', 'ownerId', 'usedby', 'status') VALUES for($i=0;$i<$npin;$i++) { ('$pin[$i]','$ownerid', 'Free', '1'); } ;"; how can I do that?

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  • MySQL - Order results by relevancy, LEFT JOINS and more

    - by XaviEsteve
    Hi guys, I am trying to get some results ordered by total votes (where client votes count 2 points and other people votes are 1 point). tab_names: +-----------+ | Name | id | +------+----+ | John | 1 | | Paul | 2 | +------+----+ tab_votes: +--------+-----------+ | idname | ip | +--------+-----------+ | 2 | 127.0.0.1 | | 2 | 127.0.0.1 | | 2 | 82.23.5.1 | | 1 | 127.0.0.1 | +--------+-----------+ This is the MySQL query I've got but doesn't work: SELECT * COUNT(v.idname) AS totalvotes, (SELECT COUNT(v.ip) FROM tab_votes WHERE v.ip LIKE '$ip') AS uservotes FROM tab_names n LEFT JOIN tab_votes v ON n.id = v.idname GROUP BY n.name ORDER BY uservotes DESC, totalvotes DESC LIMIT 40

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  • Query to MySQL from c# returns System.Byte[]

    - by karthik
    I am using the below SP to return the value of Generated Insert statement and it works fine when executed in Query browser. When i try to get the value from C#, it give's me "System.Byte[]" as return value. When i try to get the value from MySql query browser, it give's me return value as : 'insert into admindb.accounts values("54321","2","karthik2","karthik2","1");' I guess the problem is with the single quotes of the returned value. Is it so ? DELIMITER $$ DROP PROCEDURE IF EXISTS `admindb`.`InsGen` $$ CREATE DEFINER=`root`@`localhost` PROCEDURE `InsGen`( in_db varchar(20), in_table varchar(20), in_ColumnName varchar(20), in_ColumnValue varchar(20) ) BEGIN declare Whrs varchar(500); declare Sels varchar(500); declare Inserts varchar(2000); declare tablename varchar(20); declare ColName varchar(20); set tablename=in_table; # Comma separated column names - used for Select select group_concat(concat('concat(\'"\',','ifnull(',column_name,','''')',',\'"\')')) INTO @Sels from information_schema.columns where table_schema=in_db and table_name=tablename; # Comma separated column names - used for Group By select group_concat('`',column_name,'`') INTO @Whrs from information_schema.columns where table_schema=in_db and table_name=tablename; #Main Select Statement for fetching comma separated table values set @Inserts=concat("select concat('insert into ", in_db,".",tablename," values(',concat_ws(',',",@Sels,"),');') as MyColumn from ", in_db,".",tablename, " where ", in_ColumnName, " = " , in_ColumnValue, " group by ",@Whrs, ";"); PREPARE Inserts FROM @Inserts; EXECUTE Inserts; END $$ DELIMITER ;

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  • MySQL - query to return CSV in a field?

    - by StackOverflowNewbie
    Assume I have the following tables: TABLE: foo - foo_id (PK) TABLE: tag - tag_id (PK) - name TABLE: foo_tag - foo_tag_id (PK) - foo_id (FK) - tag_id (FK) How do I query this so that I get a result like this: ========================== | foo_id | tags | ========================== | 1 | foo, bar | | 2 | foo | | 3 | bar | -------------------------- Basically, I need all of foo's tags in one column, comma separated. Possible in MySQL?

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  • problem in fetching data from several tables in one query

    - by Mac Taylor
    hey guys in an attempt to union my querries into one query to database , now im in need of geting username of first poster and last poster of a topic in my forums here is my code to do as i told :: $result = $db->sql_query("SELECT t.*,p.*,u.* SUM(t.topic_approved='1') AS Amount_Of_Topics, SUM(p.post_approved ='1') AS Amount_Of_Posts FROM bb3topics t, bb3posts p, bb3users u GROUP BY t.topic_last_post_id ORDER BY t.topic_last_post_id DESC LIMIT 10 " ); while( $row = $db->sql_fetchrow($result) ) { $Amount_Of_Topics = $row['Amount_Of_Topics']; $Amount_Of_Posts = $row['Amount_Of_Posts']; $Amount_Of_Topic_Replies = $Amount_Of_Topic_Replies + $row['topic_replies']; $Amount_Of_Topic_Views = $Amount_Of_Topic_Views + $row['topic_views']; $topic_id = $row['topic_id']; $forum_id = $row['forum_id']; $topic_last_post_id = $row['topic_last_post_id']; $topic_title = $row['topic_title']; $topic_poster = $row['topic_poster']; $topic_views = $row['topic_views']; $topic_replies = $row['topic_replies']; $topic_moved_id = $row['topic_moved_id']; $topic_time = $row['topic_time']; $result2 = $db->sql_query( "SELECT topic_id, poster_id, post_time FROM bb3posts where post_id = '$topic_last_post_id'" ); list( $topic_id, $poster_id, $post_time ) = $db->sql_fetchrow( $result2 ); $result3 = $db->sql_query( "SELECT username, user_id FROM bb3users where user_id='$poster_id'" ); list( $uname, $uid ) = $db->sql_fetchrow( $result3 ); $LastPoster = "$uname"; $result4 = $db->sql_query( "SELECT username, user_id FROM bb3users where user_id='$topic_poster'" ); list( $uname, $uid ) = $db->sql_fetchrow( $result4 ); $OrigPoster = "$uname"; now i need to query all this together not in separated ones i tried using left join but didn't worked what mysql conjunction should i use ?!

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  • mysql whats wrong with this query?

    - by Hailwood
    I'm trying to write a query that selects from four tables campaignSentParent csp campaignSentEmail cse campaignSentFax csf campaignSentSms css Each of the cse, csf, and css tables are linked to the csp table by csp.id = (cse/csf/css).parentId The csp table has a column called campaignId, What I want to do is end up with rows that look like: | id | dateSent | emailsSent | faxsSent | smssSent | | 1 | 2011-02-04 | 139 | 129 | 140 | But instead I end up with a row that looks like: | 1 | 2011-02-03 | 2510340 | 2510340 | 2510340 | Here is the query I am trying SELECT csp.id id, csp.dateSent dateSent, COUNT(cse.parentId) emailsSent, COUNT(csf.parentId) faxsSent, COUNT(css.parentId) smsSent FROM campaignSentParent csp, campaignSentEmail cse, campaignSentFax csf, campaignSentSms css WHERE csp.campaignId = 1 AND csf.parentId = csp.id AND cse.parentId = csp.id AND css.parentId = csp.id; Adding GROUP BY did not help, so I am posting the create statements. csp CREATE TABLE `campaignsentparent` ( `id` int(11) NOT NULL AUTO_INCREMENT, `campaignId` int(11) NOT NULL, `dateSent` datetime NOT NULL, `account` int(11) NOT NULL, `status` varchar(15) NOT NULL DEFAULT 'Creating', PRIMARY KEY (`id`) ) ENGINE=MyISAM AUTO_INCREMENT=2 DEFAULT CHARSET=latin1 cse/csf (same structure, different names) CREATE TABLE `campaignsentemail` ( `id` int(11) NOT NULL AUTO_INCREMENT, `parentId` int(11) NOT NULL, `contactId` int(11) NOT NULL, `content` text, `subject` text, `status` varchar(15) DEFAULT 'Pending', PRIMARY KEY (`id`) ) ENGINE=MyISAM AUTO_INCREMENT=140 DEFAULT CHARSET=latin1 css CREATE TABLE `campaignsentsms` ( `id` int(11) NOT NULL AUTO_INCREMENT, `parentId` int(11) NOT NULL, `contactId` int(11) NOT NULL, `content` text, `status` varchar(15) DEFAULT 'Pending', PRIMARY KEY (`id`) ) ENGINE=MyISAM AUTO_INCREMENT=141 DEFAULT CHARSET=latin1

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  • Timeout error occurred trying to start MySQL Daemon. CentOS 5

    - by epema
    I ran into troubles with MySQL on my CentOS. I had some problems and backed up my database and removed mysql with all dependencies. After that I ran reinstalled: yum groupinstall "MySQL Database" Installed without errors. Running the mysql daemon: service mysqld start Timeout error occurred trying to start MySQL Daemon. Starting MySQL: [FAILED] I also ran # /usr/bin/mysql_install_db --user=mysql Installing MySQL system tables... 120112 1:49:44 [ERROR] Error message file '/usr/share/mysql/english/errmsg.sys' had only 480 error messages, but it should contain at least 481 error messages. Check that the above file is the right version for this program! 120112 1:49:44 [ERROR] Aborting Installation of system tables failed! Examine the logs in /var/lib/mysql for more information. You can try to start the mysqld daemon with: /usr/libexec/mysqld --skip-grant & and use the command line tool /usr/bin/mysql to connect to the mysql database and look at the grant tables: shell> /usr/bin/mysql -u root mysql mysql> show tables Try 'mysqld --help' if you have problems with paths. Using --log gives you a log in /var/lib/mysql that may be helpful. The latest information about MySQL is available on the web at http://www.mysql.com Please consult the MySQL manual section: 'Problems running mysql_install_db', and the manual section that describes problems on your OS. Another information source is the MySQL email archive. Please check all of the above before mailing us! And if you do mail us, you MUST use the /usr/bin/mysqlbug script! Checking the logs: less /var/log/mysqld.log Log file is empty. I don't even know how to debug it and not sure what to do. Any recommendations? Thank you

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