You have an error in your SQL syntax; Logging failed |
You have an error in your SQL syntax; Logging failed |
evbeej |
Jan 18 2016, 06:23 AM
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#1
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Group: Members Posts: 4 Joined: 18-January 16 Member No.: 23,917 |
Hello,
I have received the following error on two of our sites (that are on the same server) QUOTE Logging failedYou have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'get = '[]', session='[]', ip = '81.138.223.98', pageLevel = '', userLevel = '', ' at line 1 the following is where i think it is saying the problem is...: QUOTE $sessionCookieExpireTime=8*60*60; session_set_cookie_params($sessionCookieExpireTime); if (!$_SESSION) { session_start(); } if ($_GET['a'] == "logout" || $logout == "logout") { session_destroy(); } require_once("ugs/class_ugs_client.php"); $ugs = new UGS; $ugs->addLog(json_encode($_POST),json_encode($_GET),json_encode($_SESSION),$_SERVER['REMOTE_ADDR'],$page_level,json_encode($_SERVER)); any help would really be appreciated. |
CharlesEF |
Jan 18 2016, 06:46 AM
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#2
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Programming Fanatic Group: Members Posts: 1,981 Joined: 27-April 13 From: Edinburg, Texas Member No.: 19,088 |
Without seeing the addLog code from the class I don't think you have provided enough information. Are you sure you have provided the correct number of arguments for the call?
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evbeej |
Jan 18 2016, 06:50 AM
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#3
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Group: Members Posts: 4 Joined: 18-January 16 Member No.: 23,917 |
Hello, sorry where can i find the addLog code? This is the first time ive come across this sort of thing.
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evbeej |
Jan 18 2016, 07:04 AM
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#4
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Group: Members Posts: 4 Joined: 18-January 16 Member No.: 23,917 |
Ok, it looks like the following function is what is causing the error... :
QUOTE function addLog($post,$get,$sess,$ip,$pageLevel=0,$server=0) { mysql_query("INSERT INTO logs SET post='".mysql_real_escape_string($post)."', get = '".mysql_real_escape_string($get)."', session='".mysql_real_escape_string ($sess)."', ip = '{$ip}', pageLevel = '{$pageLevel}', userLevel = '{$_SESSION['role']}', userid = '{$_SESSION['uuid']}', sessionid = '".session_id()."', url = '".mysql_real_escape_string($_SERVER['REQUEST_URI'])."', server='".mysql_real_escape_string($server)."'",$this->db) or die("Logging failed".mysql_error()); } So i am going to investigate this a bit further. As it was not where i thought it was at all. |
CharlesEF |
Jan 18 2016, 07:57 AM
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#5
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Programming Fanatic Group: Members Posts: 1,981 Joined: 27-April 13 From: Edinburg, Texas Member No.: 19,088 |
Nothing jumps out at me except that you are still using mysql_* functions. mysql_* functions have been removed in PHP v7 and if you ever upgrade to that version your code will no longer work. All your code will have to be rewritten to use mysqli_* or PDO functions.
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evbeej |
Jan 18 2016, 09:59 AM
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#6
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Group: Members Posts: 4 Joined: 18-January 16 Member No.: 23,917 |
I think that may be it, all i know is i commented out that query and both sites now show the sites correctly.
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