ballhogjoni Posted July 4, 2007 Share Posted July 4, 2007 I keep getting this error: Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/realfina/public_html/thediaperbakery/left_nav.php on line 11 this is my code: mysql_connect('localhost',$username,$password); @mysql_select_db($database) or die( "Unable to select database"); $query = mysql_query('SELECT * FROM leftNavigation WHERE ID = 1'); while ($row = mysql_fetch_assoc($query)) { // line 11 $Category = $row["Category"]; } Link to comment https://forums.phpfreaks.com/topic/58428-mysql-syntax-error/ Share on other sites More sharing options...
cooldude832 Posted July 4, 2007 Share Posted July 4, 2007 if you have 0 results it will error try this and see if you have errors $query = mysql_query('SELECT * FROM leftNavigation WHERE ID = 1') or die (mysql_error()); Link to comment https://forums.phpfreaks.com/topic/58428-mysql-syntax-error/#findComment-289718 Share on other sites More sharing options...
HuggieBear Posted July 4, 2007 Share Posted July 4, 2007 Firstly I'd remove the @ from the front of @mysql_select_db. @ supresses any error messages, which if you're having problems, is probably kind of useful. Regards Huggie Link to comment https://forums.phpfreaks.com/topic/58428-mysql-syntax-error/#findComment-289721 Share on other sites More sharing options...
trq Posted July 4, 2007 Share Posted July 4, 2007 Always check your query succeeds before trying to use any result it produces. <?php mysql_connect('localhost',$username,$password); mysql_select_db($database) or die( "Unable to select database"); if ($result = mysql_query('SELECT * FROM leftNavigation WHERE ID = 1')) { if (mysql_num_rows($result)) { while ($row = mysql_fetch_assoc($result)) { $Category = $row["Category"]; } } } else { // good place to debug. echo mysql_error(); } ?> Where is $database defined? Link to comment https://forums.phpfreaks.com/topic/58428-mysql-syntax-error/#findComment-289722 Share on other sites More sharing options...
ballhogjoni Posted July 4, 2007 Author Share Posted July 4, 2007 just above all that. Thanks i got it. Link to comment https://forums.phpfreaks.com/topic/58428-mysql-syntax-error/#findComment-289728 Share on other sites More sharing options...
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