postgreSQL - 在与任何
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postgreSQL - in vs any
提问by PROvlima
I have tried both
我都试过了
1) smthng = any(select id from exmplTable)
1) smthng = 任何（从 exmplTable 中选择 id）
2) smthng in(select id from exmplTable)
2）smthng在（选择exmplTable ID）
and I am getting the same results for my data.
我的数据得到了相同的结果。
Is there any difference for the two expresions?
这两个表达式有什么区别吗？
采纳答案by Pavel Stehule
No, in these variants are same:
不，在这些变体中是相同的：
You can see - the execution plans are same too:
您可以看到 - 执行计划也相同：
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by WigglyWorld
This may be an edge case but:
这可能是一个边缘情况，但是：
select * from myTable where id IN ()
will produce: ERROR: syntax error at or near ")"
将产生：错误：“)”处或附近的语法错误
but
但
select * from myTable where id = ANY('{}');
Will return an empty result set
将返回一个空的结果集
