On Tue, Dec 28, 2010 at 5:13 AM, Jie Li <jay23j...@gmail.com> wrote: > Hi, > > Please see the following plan: > > postgres=# explain select * from small_table left outer join big_table using > (id); > QUERY PLAN > ---------------------------------------------------------------------------- > Hash Left Join (cost=126408.00..142436.98 rows=371 width=12) > Hash Cond: (small_table.id = big_table.id) > -> Seq Scan on small_table (cost=0.00..1.09 rows=9 width=8) > -> Hash (cost=59142.00..59142.00 rows=4100000 width=8) > -> Seq Scan on big_table (cost=0.00..59142.00 rows=4100000 > width=8) > (5 rows) > > Here I have a puzzle, why not choose the small table to build hash table? It > can avoid multiple batches thus save significant I/O cost, isn't it?
Yeah, you'd think. Can you post a full reproducible test case? -- Robert Haas EnterpriseDB: http://www.enterprisedb.com The Enterprise PostgreSQL Company -- Sent via pgsql-hackers mailing list (pgsql-hackers@postgresql.org) To make changes to your subscription: http://www.postgresql.org/mailpref/pgsql-hackers
