From: Masaru Sugawara <rk73@sea.plala.or.jp>
To: jack <datactrl@tpg.com.au>
Cc: pgsql-sql@postgresql.org
Subject: Re: the best way to get the topest 3 record in every group
Date: Tue, 10 Sep 2002 01:23:29 +0900
Message-ID: <20020910012302.828A.RK73@sea.plala.or.jp> (raw)
In-Reply-To: <000701c257d8$1163c9a0$1400a8c0@jac>
References: <3D7C4108.70407@mail.ru>
<000701c257d8$1163c9a0$1400a8c0@jac>
On Mon, 9 Sep 2002 18:08:21 +1000
"jack" <datactrl@tpg.com.au> wrote:
> Dima
> My question is that I want to produce ALL the lastest 3 records for EACH
> itemNo and supplier.
>
> Jack
I often use the next query for ranking, which is equivalent to Oracle's
PARTITION BY(). It isn't influenced by what kind of data your table has;
it only depends on the result of sorting
CREATE TEMP SEQUENCE seq_purchase;
SELECT t5.item_no, t5.supplier, t5.purchase_date,
t5.price, t5.qty, t5.i - t3.n + 1 AS rank
FROM (SELECT t2.item_no, t2.supplier, min(t2.i) AS n
FROM (SELECT t1.*, nextval('seq_purchase') - 1 AS i
FROM (SELECT (SELECT setval('seq_purchase',1)), *
FROM purchase
ORDER BY item_no, supplier, purchase_date desc
) AS t1
LIMIT ALL
) AS t2
GROUP BY t2.item_no, t2.supplier
) AS t3,
(SELECT t4.*, nextval('seq_purchase') - 1 AS i
FROM (SELECT (SELECT setval('seq_purchase',1)), *
FROM purchase
ORDER BY item_no, supplier, purchase_date DESC
) AS t4
LIMIT ALL
) AS t5
WHERE t3.item_no = t5.item_no
AND t3.supplier = t5.supplier
AND t5.i - t3.n + 1 <= 3
;
Regards,
Masaru Sugawara
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To: pgsql-sql@postgresql.org
Cc: rk73@sea.plala.or.jp, datactrl@tpg.com.au
Subject: Re: the best way to get the topest 3 record in every group
In-Reply-To: <20020910012302.828A.RK73@sea.plala.or.jp>
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