brem13 Posted May 31, 2009 Share Posted May 31, 2009 i keep getting this error when trying to update a table in my database - 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 'like = 1 WHERE id = 2' at line 1 code $like = htmlspecialchars($_GET['likes']); $id = htmlspecialchars($_GET['id']); $category = htmlspecialchars($_GET['category']); if($like <= 0 || $like == ""){ $like = 1; mysql_connect('*********', '*******', '***********') or die (mysql_error()); mysql_select_db('********') or die ("Could not select database because ".mysql_error()); mysql_query("UPDATE $category SET like = $like WHERE id = $id") or die("Could not insert data because ".mysql_error()); mysql_close(); }//end if ive been trying to fix this for over an hour, someone please help Link to comment https://forums.phpfreaks.com/topic/160325-solved-error-updating-table/ Share on other sites More sharing options...
phpretard Posted May 31, 2009 Share Posted May 31, 2009 Change your htmlspecialchars to striplslashes Then insert them with htmlspecialchars $like = striplslashes($_GET['likes']); update table set Like='$like' where ... Link to comment https://forums.phpfreaks.com/topic/160325-solved-error-updating-table/#findComment-846048 Share on other sites More sharing options...
brem13 Posted May 31, 2009 Author Share Posted May 31, 2009 nope, same thing Link to comment https://forums.phpfreaks.com/topic/160325-solved-error-updating-table/#findComment-846049 Share on other sites More sharing options...
phpretard Posted May 31, 2009 Share Posted May 31, 2009 Can you post $like or an example of it? Link to comment https://forums.phpfreaks.com/topic/160325-solved-error-updating-table/#findComment-846050 Share on other sites More sharing options...
phpretard Posted May 31, 2009 Share Posted May 31, 2009 ("UPDATE $category SET like = '$like' WHERE id = '$id' ") Single quotes around you variables...I didn't see that before Link to comment https://forums.phpfreaks.com/topic/160325-solved-error-updating-table/#findComment-846051 Share on other sites More sharing options...
brem13 Posted May 31, 2009 Author Share Posted May 31, 2009 like is 0 i've echoed all the variables and they all show fine this if statement should take like if like=0 and make the like field in the table = 1. i have an elseif statement underneath it, that checks if like>0 and if it is, it will add 1 to the total of like and in turn update the field in the table oh, and i've tried the single quotes, i get the same error Link to comment https://forums.phpfreaks.com/topic/160325-solved-error-updating-table/#findComment-846053 Share on other sites More sharing options...
phpretard Posted May 31, 2009 Share Posted May 31, 2009 I think like is a mysql keyword so have to use the weirdo single quotes ("UPDATE $category SET `like` = '$like' WHERE id = '$id' ") I would personly rename the DB field "like_it" or "love" Link to comment https://forums.phpfreaks.com/topic/160325-solved-error-updating-table/#findComment-846054 Share on other sites More sharing options...
brem13 Posted May 31, 2009 Author Share Posted May 31, 2009 thanks a lot, u fixed it, it was those backticks grrrrrrr, lol. thank you Link to comment https://forums.phpfreaks.com/topic/160325-solved-error-updating-table/#findComment-846057 Share on other sites More sharing options...
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