Re: 2 conditions why wont they work??

Thanks, $clientID is a string but is not empty...already tried echo($clientID) and it is not empty. Did you mean that if the value of $cientID is a string it wont work???? Thanks Brad > From: Bogdan Stancescu <mgv@fx.ro> > Date: Mon, 11 Feb 2002 05:38:31 +0200 > To: Brad Wright <bradwright@optushome.com.au> > Cc: PHP General List <php-general@lists.php.net> > Subject: Re: [PHP] 2 conditions why wont they work?? > > Do an echo($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 >