I suspect he wants:
SELECT users.uid
FROM users LEFT JOIN picks
ON users.uid = picks.user_id
WHERE picks.user_id IS NULL
Paul DuBois wrote:
> At 10:36 AM -0400 8/29/01, Jeremy Morano wrote:
>
>> Hi ,
>>
>> this is my query
>>
>> SELECT users.uid FROM users, picks WHERE users.uid = picks.user_id;
>>
>> this works correctly. The results are what they are supposed to be:
>> However, when I change the = sign to a <> or !=, The results are
>> completely
>> incorrect.
>>
>> ----------------------------------
>>
>> picks.user_id contains: 5, 1, 7, 8, 9, 12, 13, 15
>>
>> users.uid contains: 1, 8, 9, 5, 7, 10, 11, 12, 13, 14, 15
>>
>>
>> the result for SELECT users.uid FROM users, picks WHERE users.uid =
>> picks.user_id; is:
>>
>> 5, 1, 7, 8, 9, 12, 13, 15
>>
>> and the result for SELECT users.uid FROM users, picks WHERE users.uid <>
>> picks.user_id; is:
>>
>> 1, 7, 8, 9, 10, 11, 12, 13, 14, 15, 5, 7, 8, 9, 10, 11,12, 13, 14, 15,
>> 1, 5,
>> 8, 9, 10 ,11...etc
>>
>> it goes on for 80 rows with no particular pattern......H.E.L.P.!.
>
>
>
> Why do you think this is incorrect? Change the query to begin like this:
>
> SELECT users.uid, picks.user_id FROM ...
>
> and then you'll see that there is a pattern, and that the query is
> producing
> exactly what it's supposed to.
>
> I'm guessing that you're really trying to ask a different question
> than the one the query is answering. What exactly are you expecting
> as output?
--
Gerald L. Clark
[EMAIL PROTECTED]
---------------------------------------------------------------------
Before posting, please check:
http://www.mysql.com/manual.php (the manual)
http://lists.mysql.com/ (the list archive)
To request this thread, e-mail <[EMAIL PROTECTED]>
To unsubscribe, e-mail <[EMAIL PROTECTED]>
Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php
