[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
[ale] any SQL gurus?
- Subject: [ale] any SQL gurus?
- From: matthew.brown at cordata.com (Matthew Brown)
- Date: Wed Jan 21 11:19:23 2004
- In-reply-to: <[email protected]>
Nahh.. No need for a sub-select on this one. Lemme ponder.
(888) 681-2262 . (770) 795-0089 .
* Windows and Linux Networks
* Web Development & Hosting
* Application Development & Hosting
From: ale-bounces at ale.org [mailto:ale-bounces at ale.org] On Behalf Of Jason
Sent: Wednesday, January 21, 2004 11:04 AM
To: Atlanta Linux Enthusiasts
Subject: Re: [ale] any SQL gurus?
On Jan 21, 2004, at 10:39 AM, Jeff Tillotson wrote:
> Not a SQL guru but wouldn't this work:
> SELECT COPY.owner COPY.title_key TITLE.name from TITLE COPY where
> TITLE.title_key = COPY.title_key order by COPY.owner;
On Jan 21, 2004, at 10:34 AM, Dylan Northrup wrote:
> I'm not much of a SQL guru, but wouldn't this work:
> select t.name from t TITLE, c COPY where c.owner = "whoever" and
> c.title_key = f.title_key;
Thanks guys. Hmm. Both of these work if I use the DISTINCT keyword.
But for some reason, such things as DISTINCT COUNT(name) don't work, and
each copy goes toward the total. I can't really afford to fetch all the
results and count them on the client side. I shouldn't have used the word
"list", my mistake.
Hmm. Looks like the inner join does the same thing--or were these other
queries already implicitly inner joins? I really dislike databases. :-)
So it's not a list I need, but a count. Hrmm.
I bet I need sub-selects. Something like SELECT COUNT(*) FROM (
SELECT DISTINCT )
Thanks for your help. I'm going to keep banging my head against it.
Ale mailing list
Ale at ale.org