This code works and enters all of the correct information into the database but the error check returns an error. I could remove the error check but I'm afraid of creating some nightmare that comes back to haunt me later or that I'm missing a fundamental issue:

```
    $sql5a = mysql_query("SELECT id FROM categories WHERE category='$category'");
        $categoryresult = mysql_fetch_array($sql5a);
        $oldcategoryid = $categoryresult['id'];
        $sql6a = "INSERT INTO lookupcategory SET
        fbid='$fbid2',
        categoryid='$oldcategoryid'";
            if ( @mysql_query($sql5b) ) { 
                                echo('sql5b updated successfully<br>'); 
                            } else { 
                                echo('Error: sql5b not updated<br>'.mysql_error() );
                        }
                            if ( @mysql_query($sql6b) ) { 
                                echo('sql6b updated successfully<br>'); 
                            } else { 
                                echo('Error: sql6b not updated<br>'.mysql_error() );
                        }

```

The output is: "Error: sql5b not updated
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Resource id #7' at line 1"
"sql6b updated successfully"

When I check the database all entries are correct. If sql5a didn't work, sql6b couldn't work, hence my confusion over the error.

A sample Category would be: Travel/Leisure
The category was originally created from a form response:
$category = htmlentities($fbdetail['category'], ENT\_QUOTES); 
and entered into the database successfully. An id number was assigned using AUTO\_INCREMENT.