On Wed, 22 Jan 2003, Tomasz Myrta wrote:
> Let's make some test:
>
> First, let's create some simple view with 2 tables join:
> drop view pm;
> create view pm as
> select
> id_przystanku,
> m.nazwa
> from
> przystanki p
> join miasta m using (id_miasta);
>
> explain select * from pm where id_przystanku=1230;
> Nested Loop (cost=0.00..6.26 rows=1 width=23)
> -> Index Scan using przystanki_pkey on przystanki p (cost=0.00..3.14 rows=1 width=8)
> -> Index Scan using miasta_pkey on miasta m (cost=0.00..3.10 rows=1 width=15)
>
>
> Next, let's try query using this view 2 times with explicit join:
> explain select * from pm a join pm b using(id_przystanku) where id_przystanku=1230;
> Hash Join (cost=13.00..30.10 rows=1 width=46)
> -> Hash Join (cost=6.74..21.02 rows=374 width=23)
> -> Seq Scan on przystanki p (cost=0.00..7.74 rows=374 width=8)
> -> Hash (cost=5.99..5.99 rows=299 width=15)
> -> Seq Scan on miasta m (cost=0.00..5.99 rows=299 width=15)
> -> Hash (cost=6.26..6.26 rows=1 width=23)
> -> Nested Loop (cost=0.00..6.26 rows=1 width=23)
> -> Index Scan using przystanki_pkey on przystanki p (cost=0.00..3.14 rows=1 width=8)
> -> Index Scan using miasta_pkey on miasta m (cost=0.00..3.10 rows=1 width=15)
>
> And now similiar view, but without nesting views:
> drop view pm2;
> create view pm2 as
> select
> id_przystanku,
> m1.nazwa as nazwa1,
> m2.nazwa as nazwa2
> from
> przystanki p1
> join miasta m1 using (id_miasta)
> join przystanki p2 using (id_przystanku)
> join miasta m2 on (m2.id_miasta=p2.id_miasta);
That's not the same join for optimization purposes
since postgresql treats explicit join syntax as a
constraint on the ordering of joins.
The same join would be something like:
przystanki p1 join miasta m1 using (id_miasta)
join (przystanki p2 join miasta m2 using (id_miasta))using (id_przystanku)
minus the fact I think you'd need some explicit naming in
there.
