问题
I have tried both
1) smthng = any(select id from exmplTable)
2) smthng in (select id from exmplTable)
and I am getting the same results for my data.
Is there any difference for the two expresions?
回答1:
No, in these variants are same:
You can see - the execution plans are same too:
postgres=# explain select * from foo1 where id in (select id from foo2); ┌──────────────────────────────────────────────────────────────────┐ │ QUERY PLAN │ ╞══════════════════════════════════════════════════════════════════╡ │ Hash Semi Join (cost=3.25..21.99 rows=100 width=4) │ │ Hash Cond: (foo1.id = foo2.id) │ │ -> Seq Scan on foo1 (cost=0.00..15.00 rows=1000 width=4) │ │ -> Hash (cost=2.00..2.00 rows=100 width=4) │ │ -> Seq Scan on foo2 (cost=0.00..2.00 rows=100 width=4) │ └──────────────────────────────────────────────────────────────────┘ (5 rows) postgres=# explain select * from foo1 where id = any (select id from foo2); ┌──────────────────────────────────────────────────────────────────┐ │ QUERY PLAN │ ╞══════════════════════════════════════════════════════════════════╡ │ Hash Semi Join (cost=3.25..21.99 rows=100 width=4) │ │ Hash Cond: (foo1.id = foo2.id) │ │ -> Seq Scan on foo1 (cost=0.00..15.00 rows=1000 width=4) │ │ -> Hash (cost=2.00..2.00 rows=100 width=4) │ │ -> Seq Scan on foo2 (cost=0.00..2.00 rows=100 width=4) │ └──────────────────────────────────────────────────────────────────┘ (5 rows)
回答2:
This may be an edge case but:
select * from myTable where id IN ()
will produce: ERROR: syntax error at or near ")"
but
select * from myTable where id = ANY('{}');
Will return an empty result set
来源:https://stackoverflow.com/questions/30263671/postgresql-in-vs-any