forumnz Posted January 3, 2008 Share Posted January 3, 2008 I have modified my scrip to include the variable 'region', but now it won't work. Heres the code snippet I modified: $query = ("SELECT * FROM cmads WHERE AND (keyw LIKE '%$trimmed%' OR buzzname LIKE '%$trimmed%' OR descr LIKE '%$trimmed%' OR sub LIKE '%$trimmed%') AND (region LIKE '$region') ORDER BY keyw"); $numresults=mysql_query($query); $numrows=mysql_num_rows($numresults); All I did was add the AND (region LIKE '$region') And my error: Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /usr/local/www/vhosts/aaa/httpdocs/aaa/Results.php on line 84 Line 84 is: $numrows=mysql_num_rows($numresults); Any idea? Thanks, Sam. Link to comment https://forums.phpfreaks.com/topic/84344-solved-warning-mysql_num_rows-error/ Share on other sites More sharing options...
GingerRobot Posted January 3, 2008 Share Posted January 3, 2008 There appears to be a stray 'AND' in there. Try: $query = "SELECT * FROM cmads WHERE (keyw LIKE '%$trimmed%' OR buzzname LIKE '%$trimmed%' OR descr LIKE '%$trimmed%' OR sub LIKE '%$trimmed%') AND (region LIKE '$region') ORDER BY keyw"; $numresults=mysql_query($query) or die(mysql_error().'<br />Query'.$query); $numrows=mysql_num_rows($numresults); I've also added an or die statement to the query - it'll show us the mysql error if there is one. Link to comment https://forums.phpfreaks.com/topic/84344-solved-warning-mysql_num_rows-error/#findComment-429604 Share on other sites More sharing options...
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