When you get that error, it generally means your query failed, or you have the wrong variable in a mysql_fetch_*() function. Use mysql_error() to see what the error is if the query failed. ---John W. Holmes... PHP Architect - A monthly magazine for PHP Professionals. Get your copy today. http://www.phparch.com/ > -----Original Message----- > From: Addison Ellis [mailto:addison@bellsouth.net] > Sent: Monday, January 20, 2003 2:30 PM > To: php-db@lists.php.net > Subject: not a valid MySQL result resource > > hello and thank you for your time. > can you tell me why the below: i think this whole section is > line 22 because whatever i change in these first four lines still has > the error message point to line 22. > $cobj = mysql_db_query($dbname,"select * from category where > id=$category"); //line 22 > $crow = mysql_fetch_object($cobj); > $scobj = mysql_db_query($dbname,"select * from subcategory where > category=$category"); > $scrow = mysql_fetch_object($scobj); > > <? echo $crow->name;?> > <? echo $scrow->name;?> > > gives me this: > Warning: Supplied argument is not a valid MySQL result resource in > /users/infoserv/web/register/ca/direct.php on line 22 > > > perhaps i should use something different. i am trying to print the > results from two previous selections on this third page. the second > page works with > $cobj = mysql_db_query($dbname,"select * from category where > id=$category"); > $crow = mysql_fetch_object($cobj); > <? echo $crow->name;?> > > and then they make selection two... > thank you again. addison > -- > Addison Ellis > small independent publishing co. > 114 B 29th Avenue North > Nashville, TN 37203 > (615) 321-1791 > addison@bellsouth.net > info@smipco.com > subsidiaries of small independent publishing co. > info@gloabaldog.com > info@momandpocentral.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
