-----Original Message-----
From: pgsql-general-owner@postgresql.org
[mailto:pgsql-general-owner@postgresql.org] On Behalf Of Nick
Sent: Tuesday, January 17, 2012 12:59 PM
To: pgsql-general@postgresql.org
Subject: Re: [GENERAL] Whats the most efficient query for this result?
On Jan 17, 3:33 am, t...@audioboundary.com (Tom Molesworth) wrote:
> Hi Nick,
>
> On 17/01/12 00:18, Nick wrote:
>
> > I have three tables (users, books, pencils) and would like to get a
> > list of all users with a count and total price of their books and
> > pencils for 2012-01-01...
>
> > So with this data...
>
> > users (user_id)
> > 1
> > 2
> > 3
>
> > books (user_id, price, created)
> > 1 | $10 | 2012-01-01
> > 1 | $10 | 2012-01-01
> > 3 | $10 | 2012-01-01
>
> > pencils
> > 1 | $.50 | 2012-01-02
> > 3 | $.50 | 2012-01-01
> > 3 | $.50 | 2012-01-02
>
> > What is the most efficient way to get this result...
>
> > query_result (user_id, book_count, book_price_total, pencil_count,
> > pencil_price_total)
> > 1 | 2 | $20 | 0 | $0
> > 2 | 0 | $0 | 0 | $0
> > 3 | 1 | $10 | 1 | $.50
>
> Seems straightforward enough - left join the tables, group the result
> on user_id - so I'd write it as:
>
> select u.user_id, count(b.user_id) as "book_count",
> coalesce(sum(b.price), 0) as "book_price_total", count(p.user_id) as
> "pencil_count", coalesce(sum(b.price), 0) as "pencil_price_total"
> from users u
> left join books b on b.user_id = u.user_id and b.created = '2012-01-01'
> left join pencils p on p.user_id = u.user_id and p.created = '2012-01-01'
> group by u.user_id
> order by u.user_id
>
> If you need something more efficient, summary tables may help - hard
> to say without knowing more about the real data.
>
> Tom
>
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I take that back, now im incorrectly not getting user 2's results
---------------------------------------------------------------------
Not Tested & Psuedo Code But...
SELECT user_id, COALESCE(books_agg.count,0), COALESCE(books_agg.sum,0.00),
... [same for pencils]
FROM user u
LEFT JOIN ( SELECT user_id, COUNT(*), SUM(price) FROM book GROUP BY user_id
) books_agg ON ( books_agg.user_id = u.user_id )
LEFT JOIN ( SELECT user_id, COUNT(*), SUM(price) FROM pencils GROUP BY
user_id ) pencils_agg USING ( pencils_agg.user_id = u.user_id )
-- NOTE: THERE IS NO GROUP BY IN THIS PART OF THE QUERY; only in the
sub-queries
Basically do all your INDEPENDENT calculations first then simply JOIN the
various results to each other while replacing missing JOINs with reasonable
default values.
David J.