Question

If I use a variable in a php SQL statement that also has COUNT, I get an error. If use a literal number instead of the variable, it works fine. In the code below, you can see where I set my variable. It’s set to “test2”, first line of code below.

What am I doing wrong?

//$tag_text_ipb hardcoded here for testing    
$tag_text_ipb="test2";  

//when I replace $tag_text_ipb with the literal 'test2' in SQL below, it works fine.
$query_total_tags = "SELECT COUNT(1) FROM core_tags WHERE tag_meta_app = 'downloads' AND tag_text = $tag_text_ipb"; 

$dlresult_total_tags = mysql_query( $query_total_tags );

//Mysql reports an error here (see below for the error text) ONLY when I use the $tag_text_ipb variable in the SQL statement.
$tag_count= mysql_result($dlresult_total_tags,$k[COUNT(1)]);  

The error is:
Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /home/ipboard/admin/… : eval()’d code on line 3884

This error disappears and everything works properly if I use a literal in the SQL statement instead of $tag_text_ipb.

Answer 1

You’re missing quotes around the tag text var in your SQL. Should be like this:

$query_total_tags = "SELECT COUNT(1) FROM core_tags WHERE tag_meta_app = 'downloads' AND tag_text = '".$tag_text_ipb."'";