In-Reply-To: <4CF8CC26.70206@megafon.hr>
References: <4CF8CC26.70206@megafon.hr>
To: Mario Splivalo <mario.splivalo@megafon.hr>
Cc: pgsql-sql@postgresql.org,
	pgsql-sql-owner@postgresql.org
MIME-Version: 1.0
Subject: Re: Get the max viewd product_id for user_id
Message-ID: 
 <OF16FA4770.DCF7942F-ON652577EE.003F9FF3-652577EE.0040227D@ibsplc.com>
From: Jayadevan M <Jayadevan.Maymala@ibsplc.com>
Date: Fri, 3 Dec 2010 17:10:31 +0530
Content-Type: text/plain; charset="US-ASCII"

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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