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.