Syntax error?

I'm hoping this is a case of a simple syntax error and someone can clear this up fairly quickly. If more info is needed, please let me know. I'm sure I'm missing something fairly basic here, and if so, please straighten me out. I'm using this bit of code to do a simple query based on a couple of variables: " $db = mysql_connect("***server****", "****username****", "****password****"); mysql_select_db("gu_mgmt",$db); if ($subcats == "All") { $result = mysql_query("SELECT * FROM $catty",$db); } else { $result = mysql_query("SELECT * FROM $catty WHERE SubCategory = '$subcats' order by DateAcquired",$db); } " The first bit of the "if" works just fine, but the second bit seems to return an empty result. I get no errors whatsoever, I just don't get any info on the page. Does it have something to do with the way I've used the variable to represent the table name? I'm terribly confused. This script worked FINE when I was only passing the "subcats" variable. It didn't break until I tried throwing in the "catty" variable. I had the name of the table hardcoded before, but now I want to dynamically use the same script to check a few different tables in the same db. Thanks in advance for any and all help. Bradley