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

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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

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