Mysql Query Failed With Error 1064
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site. Subscribe HomeSupportSite restoration#13544: MySQL query failed with error 1064 #13544 – MySQL query failed with error 1064 Posted in ‘Site restoration’ This is a public ticket. Everybody will be able
You Have An Error In Your Sql Syntax Check The Manual That Corresponds To Your Mysql Server Version
to see its contents. Do not include usernames, passwords or any other sensitive information. Friday, 14 September 2012 15:15 CDT bcroft Mandatory information about my setup: Have I read the related troubleshooter articles above before posting (which pages?)? Yes Have I searched the tickets before posting? Yes Have I read the documentation before posting (which pages?)? Yes Joomla! version: 1.5.26 PHP version: #1064 - you have an error in your sql syntax; 5.2.5 MySQL version:5.5.25a Host: siteground Akeeba Backup version which took the backup: 3.2.7 Kickstart version used to extract the backup: 3.2 Description of my issue: I backup up my site and then installed a recommended upgrade. Plugin manager then failed and siteground restored the database from softaculous. This didn't solve so i thought it best to restore from akeeba. Kickstart works fine and then I get the error: Error at the line 0: CREATE TABLE `#__ak_profiles` ( `id` int(10) unsigned NOT NULL AUTO_INCREMENT, `description` varchar(255) NOT NULL, `configuration` longtext, `filters` longtext, PRIMARY KEY (`id`) ) TYPE=My ...MySQL: MySQL query failed with error 1064 (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 'TYPE=MyISAM AUTO_INCREMENT=2' at line 1). The query was:CREATE TABLE `jos_ak_profiles` ( `id` int(10) unsigned NOT NULL AUTO_INCREMENT, `description` varchar(255) NOT NULL, `configuration` longtext, `filters` longtext, PRIMARY KEY (`id`) ) TYPE=MyISAM AUTO_INCREMENT=2 Have tried deleting the database and recreating and it still fails. I'm not the most technical when in comes to web sites and w
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here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this http://stackoverflow.com/questions/17618665/error-received-from-database-was-1064-php-mysql site About Us Learn more about Stack Overflow the company Business Learn more http://stackoverflow.com/questions/39488348/mysql-query-failed-to-run-mysql-error-1064 about hiring developers or posting ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 6.2 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up Error received from error 1064 database was #1064 - PHP & MySQL up vote 0 down vote favorite I have a simple MySQL query that I use in PHP but it gives me this error; Database query failed. Error received from database was #1064: 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 you have an 'NULL' at line 3 for the query: SELECT * FROM mantis_user_table WHERE email LIKE '%me%'. $t_user_table = db_get_table( 'mantis_user_table' ); $temp_mail = '%' . $p_email . '%'; $query = "SELECT * FROM $t_user_table WHERE email LIKE '{$temp_mail}'"; $result = db_query_bound( $query, Array( $p_realname ) ); The thing I can not understand is, when I run it in phpmyadmin, it gives me the correct values. So can you please help me with this situation ? Thank you EDIT : This is the code that I use while editing : $File = "c:/YourFile.txt"; $Handle = fopen($File, 'w'); $Data = $query; fwrite($Handle, $Data); fclose($Handle); Here you can see the query that I get when I write : SELECT * FROM mantis_user_table WHERE email LIKE '%meh%' And this query normally gives me some results in phpmyadmin , now I dont have any result on web site And if you want this is the rest of the code that returns the id of the selected item: $result = db_query_bound( $query, Array( $temp_mail ) ); if( 0 == db_num_rows( $result ) ) { return false; } else { $row = db_fetch_array( $result ); user_cache_database_result( $row ); return
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company Business Learn more about hiring developers or posting ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 6.2 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up MySQL query failed to run - MySQL error: #1064 up vote 0 down vote favorite I'm getting the error: #1064 - 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 ' dag = zaterdag, start = 15, start_minuten = 00, eind = 19, eind_minuten = 00, d' at line 1 when I try to run this query in Phpmyadmin: SELECT * FROM rooster WHERE medewerker = 00126311, dag = zaterdag, start = 15, start_minuten = 00, eind = 19, eind_minuten = 00, datum = 2016-09-17, weeknummer = 37, leiding = 0 My table (rooster) structure: id int(4) medewerker varchar(10) dag varchar(10) start varchar(25) start_minuten varchar(25) eind varchar(25) eind_minuten varchar(25) datum date weeknummer int(2) leiding int(1) medewerker is inner joined with another table, it shows the employee's number. translation: zaterdag means saturday medewerker means employee dag means day minuten means minutes datum means date weeknummer mean weeknumber leiding means boss (if boss is 1, he's the boss on that day) I will appreciate it if someone could help me with this problem. mysql sql share|improve this question edited Sep 14 at 10:47 Shaggy 349210 asked Sep 14 at 10:38 TripleDeal 337 Use AND (or OR) between the WHERE clauses conditions, instead of commas! –jarlh Sep 14 at 10:39 But then it says, unknown column 'zaterdag' in 'where clause'. I typed this: SELECT * FROM rooster WHERE medewerker = 00126311 AND dag = zaterdag AND start = 15 AND s