Alright, PHP is throwing this error at me (in the log) when I run the code mentioned below:
mysql_num_rows() expects parameter 1 to be resource, string given in (place) on line 10
$queryFP = ("SELECT * FROM db"); $countFP = mysql_num_rows($queryFP); $aID = rand(1, $countFP);
I think it has something to do with the $queryFP's syntax, but I'm not completely sure how to fix it since $queryFP's syntax is the simplest query I've ever seen.
You need to query the database first.