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:[EMAIL PROTECTED]]
> Sent: Monday, January 20, 2003 2:30 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] 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
> [EMAIL PROTECTED]
> [EMAIL PROTECTED]
> subsidiaries of small independent publishing co.
> [EMAIL PROTECTED]
> [EMAIL PROTECTED]



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