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Re: Whats the most efficient query for this result?

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On Jan 17, 3:33 am, t...@xxxxxxxxxxxxxxxxx (Tom Molesworth) wrote:
> Hi Nick,
>
> On 17/01/12 00:18, Nick wrote:
>
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>
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> > 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

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