Yes - that's most likely the answer.. if the echo doesn't help, try something like this..
$result2 = mysql_query($query2,$db) or die(mysql_error()); If mysql is throwing up an error, that'll present it too you and stop execution.. -----Original Message----- From: Bogdan Stancescu [mailto:[EMAIL PROTECTED]] Sent: 11 February 2002 03:39 To: Brad Wright Cc: PHP General List Subject: Re: [PHP] 2 conditions why wont they work?? Do an echo($clientID) before - it most probably is either empty or it's a string and MySQL actually issues errors there. Bogdan Brad Wright wrote: > Hi all, > Im was sure you could select a row from a mySQL database based on 2 > conditions. My code: > > $query2 = "select * from Table where userNo = $userNo and clientID = > $clientID"; > $result2 = mysql_query($query2,$db); > > This returns : > Warning: Supplied argument is not a valid MySQL result resource > > b ut if i change the line to: > $query2 = "select * from Table where userNo = $userNo; > $result2 = mysql_query($query2,$db); > > it works (but get all the rows with userNo = $userNo. > > HELP!!!! Im sure this should work...what am i doing wrong??? > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
