Re: Making a Variable from different tables with Matching Dbfields?

Well upon looking it was number 2, I had no orders in the workorder table, but here is an oddity I wonder if someone can explain if I run this simple query: $query = "SELECT * FROM admin, workorders WHERE admin.UserName = '".$_SESSION['user']."' "; It returns adminID = 7 (my number is 20) username= tmiller (this is correct) it's the only query I can get to return anything but it's the wrong adminID can someone explain what can be making that happen On Thu, Jan 29, 2009 at 2:48 PM, Shawn McKenzie <nospam@mckenzies.net>wrote: > Terion Miller wrote: > > Hi Again > > > > Here is the query and code I tried: > > $sql = "SELECT * FROM workorders WHERE AdminID = (SELECT AdminID FROM > > admin WHERE UserName = '" > > . mysql_real_escape_string($_SESSION['user']) . "')"; > > > > > > $result2 = mysql_query ($sql); > > $row2 = mysql_fetch_assoc ($result2); > > $printrow = print_r($row2); > > > > > > > > Here is my print variables--- Nothing printed with the print_r: > > > > $sql "SELECT * FROM workorders WHERE AdminID = (SELECT AdminID FROM > > admin WHERE UserName = 'tmiller')" > > $result2 "Resource id #6" > > $row2 "" > > $printrow "1" > > > O.K., easy: > > 1. There is no UserName in admin that = $_SESSION['user'] > 2. Or, the AdminID that is returned for $_SESSION['user'] is not present > in workorders > > That's why I suggested the 2 query approach the first time so that you > could echo out the AdminID after the first query and then go into the db > and make sure it exists in workorders. Or, if none were returned, then > echo out $_SESSION['user'] and make sure it exists in admin. > > -Shawn >