Re: Why hash join cost calculation need reduction

From: Stephen Frost <sfrost(at)snowman(dot)net>
To: 高健 <luckyjackgao(at)gmail(dot)com>
Cc: Tom Lane <tgl(at)sss(dot)pgh(dot)pa(dot)us>, pgsql-general <pgsql-general(at)postgresql(dot)org>
Subject: Re: Why hash join cost calculation need reduction
Date: 2013-06-14 11:58:24
Message-ID: 20130614115824.GA6417@tamriel.snowman.net
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* 高健 (luckyjackgao(at)gmail(dot)com) wrote:
[...]
> postgres=# explain analyze select * from sales s inner join customers c on
> s.cust_id = c.cust_id and c.cust_id =2;
[...]
> When I use the where condition such as <cust_id=2>,
>
> postgresql is clever enough to know it is better to make seqscan and
> filter ?

I havn't bothered to go look through the code specifics, but what I
expect is happening here is that PG realizes that c.cust_id and
s.cust_id are the same, and c.cust_id = 2, therefore the statement is
equivilant to:

explain analyze select * from sales s inner join customers c on
s.cust_id = 2 and c.cust_id = 2

and it's not going to try and build a hash to support an equality
operation against a constant- there's no point. It can simply do a
nested loop with a filter because all it needs to do is find all cases
of "sales.cust_id = 2" and all cases of "customers.cust_id = 2" and
return the cartesian product of them.

Thanks,

Stephen

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