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From: Ireneusz Pluta <ipluta@wp.pl>
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To: jamcito <jamcito@poczta.fm>
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Subject: Re: partition text/varchar check problem
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jamcito napisał(a):
> /*********************************************/
> SET constraint_exclusion=on;
> SET
>
> SHOW constraint_exclusion;
>  constraint_exclusion
> ----------------------
>  on
> (1 row)
>
> EXPLAIN SELECT * FROM data WHERE name = 'aaa';
>                                QUERY PLAN
> ------------------------------------------------------------------------
>  Result  (cost=0.00..24.42 rows=7 width=48)
>    ->  Append  (cost=0.00..24.42 rows=7 width=48)
>          ->  Seq Scan on data  (cost=0.00..22.38 rows=5 width=48)
>                Filter: ((name)::text = 'aaa'::text)
>          ->  Seq Scan on data_a data  (cost=0.00..1.02 rows=1 width=23)
>                Filter: ((name)::text = 'aaa'::text)
>          ->  Seq Scan on data_b data  (cost=0.00..1.02 rows=1 width=23)
>                Filter: ((name)::text = 'aaa'::text)
> (8 rows)
>   
Can you show what you get from:
EXPLAIN SELECT * FROM data WHERE name LIKE 'a%'

?

Irek.