RE: please help .........very urgent

2005-09-07 Thread Edwin Cruz
Hi!! $dbQuery = "select * from table where 1" ; $result = mysql_query($dbQuery) or die("Error trying to execute the query: ".mysql_error()); If(mysql_num_rows($result)<1){ //also: "==0" works echo "NULL"; -Original Message- From: Kane Wilson [mailto:[EMAIL PROTECTED] Sent: Wednes

Re: please help .........very urgent

2005-09-07 Thread Gleb Paharenko
Hello. > $dbQuery = results; Have you forgotten to put '$' before the 'results'? Do you have display_errors = On in your php.ini? Kane Wilson <[EMAIL PROTECTED]> wrote: > > > hi , > > I wanted to check the following condition and if it is > success i wanted to display a massag

Re: please help .........very urgent

2005-09-06 Thread Jigal van Hemert
Kane Wilson wrote: I wanted to check the following condition and if it is success i wanted to display a massage.I tried as follows. but no luck. nothing displayed. $dbQuery = results; $result = mysql_query($dbQuery) or die("Couldn't get file list"); if (!isset($result)) {echo "NULL";}