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

Shawn McKenzie wrote: > Terion Miller wrote: >> 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 >> > > You cannot return related values from 2 tables without joining the > tables in the query which we've shown is not going to work because you > need to get AdminID from admin before you can join it to workorders. > > What your query says is, "give me all fields for all rows from admin, > where UserName = the session user, and give me the same number of rows > from workorders arbitrarily or just the first however many rows". I > would be willing to bet that the first record in workorders has AdminID = 7. > > This works, obviously: > > $query = "SELECT * FROM admin WHERE admin.UserName = > '".$_SESSION['user']."' "; > I also bet that now that you know there are no valid records for your query in the workorders table, that the join queries that others have provided will work great! -- Thanks! -Shawn http://www.spidean.com