From: Jayadevan M <Jayadevan.Maymala@ibsplc.com>
To: Mario Splivalo <mario.splivalo@megafon.hr>
Cc: pgsql-sql@postgresql.org
Cc: pgsql-sql-owner@postgresql.org
Subject: Re: Get the max viewd product_id for user_id
Date: Fri, 3 Dec 2010 17:10:31 +0530
Message-ID: <OF16FA4770.DCF7942F-ON652577EE.003F9FF3-652577EE.0040227D@ibsplc.com> (raw)
In-Reply-To: <4CF8CC26.70206@megafon.hr>
References: <4CF8CC26.70206@megafon.hr>
Hello,
> I went this way, but for a large number of user_id's, it's quite slow:
>
> CREATE VIEW v_views AS
> SELECT user_id, product_id, count(*) as views
> FROM viewlog
> GROUP BY user_id, product_id
>
> SELECT
> DISTINCT user_id,
> (SELECT product_id FROM v_views inn WHERE inn.user_id = out.user_id
> ORDER BY views DESC LIMIT 1) as product_id,
> (SELECT views FROM v_views inn WHERE inn.user_id = out.user_id ORDER
BY
> views DESC LIMIT 1) as views
> FROM
> v_views out
>
Does this work faster?
select x.user_id,y.product_id,x.count from
(select user_id, max(count ) as count from (select user_id,product_id,
count(*) as count from viewlog group by user_id,product_id) as x group by
user_id
) as x inner join
(select user_id,product_id, count(*) as count1 from viewlog group by
user_id,product_id ) as y
on x.user_id=y.user_id and x.count=y.count1
Regards,
Jayadevan
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To: pgsql-sql@postgresql.org
Cc: Jayadevan.Maymala@ibsplc.com, mario.splivalo@megafon.hr, pgsql-sql-owner@postgresql.org
Subject: Re: Get the max viewd product_id for user_id
In-Reply-To: <OF16FA4770.DCF7942F-ON652577EE.003F9FF3-652577EE.0040227D@ibsplc.com>
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