Question

This query is giving me syntax errors when I atempt to execute it in PHP mysqli, but not when executed in the MySQL CLI. Can someone tell me what is happening here.

Here’s the query:

DROP TABLE IF EXISTS `wp_commentmeta`;
    CREATE TABLE `wp_commentmeta` (
      `meta_id` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
      `comment_id` bigint(20) unsigned NOT NULL DEFAULT '0',
      `meta_key` varchar(255) DEFAULT NULL,
      `meta_value` longtext,
      PRIMARY KEY (`meta_id`),
      KEY `comment_id` (`comment_id`),
      KEY `meta_key` (`meta_key`)
    ) ENGINE=MyISAM DEFAULT CHARSET=latin1;

Here’s the test code:

<?php

$sql="
    DROP TABLE IF EXISTS `wp_commentmeta`;
    CREATE TABLE `wp_commentmeta` (
      `meta_id` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
      `comment_id` bigint(20) unsigned NOT NULL DEFAULT '0',
      `meta_key` varchar(255) DEFAULT NULL,
      `meta_value` longtext,
      PRIMARY KEY (`meta_id`),
      KEY `comment_id` (`comment_id`),
      KEY `meta_key` (`meta_key`)
    ) ENGINE=MyISAM DEFAULT CHARSET=latin1;
";

$conn=mysqli_connect('localhost','root','yesthereis','test');
if(mysqli_query($conn, $sql)){
    echo "Inserted\n";
}else{
    echo "Failed\n".mysqli_error($conn)."\n";
}
?>

…and it’s execution:

jgalley@jgalley-debian:~/code/$ php test.php 
Failed
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 'CREATE TABLE `wp_commentmeta` (
      `meta_id` bigint(20) unsigned NOT NULL AUT' at line 2

As you can see, it works fine in the CLI:

jgalley@jgalley-debian:~/code/mysqlsync2$ mysql -u root -p'yesthereis' test
Reading table information for completion of table and column names
You can turn off this feature to get a quicker startup with -A

Welcome to the MySQL monitor.  Commands end with ; or \g.
Your MySQL connection id is 213
Server version: 5.1.63-0+squeeze1 (Debian)

Copyright (c) 2000, 2011, Oracle and/or its affiliates. All rights reserved.

Oracle is a registered trademark of Oracle Corporation and/or its
affiliates. Other names may be trademarks of their respective
owners.

Type 'help;' or '\h' for help. Type '\c' to clear the current input statement.

mysql> DROP TABLE IF EXISTS `wp_commentmeta`;
Query OK, 0 rows affected, 1 warning (0.05 sec)

mysql>     CREATE TABLE `wp_commentmeta` (
    ->       `meta_id` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
    ->       `comment_id` bigint(20) unsigned NOT NULL DEFAULT '0',
    ->       `meta_key` varchar(255) DEFAULT NULL,
    ->       `meta_value` longtext,
    ->       PRIMARY KEY (`meta_id`),
    ->       KEY `comment_id` (`comment_id`),
    ->       KEY `meta_key` (`meta_key`)
    ->     ) ENGINE=MyISAM DEFAULT CHARSET=latin1;
Query OK, 0 rows affected (0.01 sec)

mysql> quit
Bye

Answer 1

You cannot run two queries in a single mysqli_query call. Break your query string in two and execute them separately. (Or, as TheVedge correctly points out in the comments, use mysqli_multi_query as a drop-in replacement for mysqli_query in your script.)

This helps mitigate SQL injections.