Question

如何让MySQL首先运行子查询？现在MySQL运行表t1中每一行的内部查询，这是一场性能灾难。

explain select * from t1 where uid in (select id from t0);
+----+--------------------+-------+------+---------------+------+---------+------+----------+-------------+
| id | select_type        | table | type | possible_keys | key  | key_len | ref  | rows     | Extra       |
+----+--------------------+-------+------+---------------+------+---------+------+----------+-------------+
|  1 | PRIMARY            | t1    | ALL  | NULL          | NULL | NULL    | NULL | 18954249 | Using where |
|  2 | DEPENDENT SUBQUERY | t0    | ALL  | NULL          | NULL | NULL    | NULL |    12749 | Using where |
+----+--------------------+-------+------+---------------+------+---------+------+----------+-------------+

Answer 1

如果您执行inner join，您将获得所需的结果。

自：

select * from t1 where uid in (select id from t0);

要：

select * from t1 join t0 on t1.t0id = t0.id

http://en.wikipedia.org/wiki/Join_(SQL)#Inner_join

Answer 2

以下2将比您当前的query工作得更快，其中一个更好地扩展取决于您的数据库结构

SELECT
  * 
FROM
  t1 t
  JOIN t0 ON ( t.uid = t0.id )      << If there are 2 row matches on table `t0`, it shall return duplicated rows (t1 table contents duplicated, you could use a distinct to avoid such cases) OR use the second query instead

和

SELECT * FROM t1 t WHERE EXISTS (select 1 from t0 WHERE t0.id = t.uid)

如何首先运行子查询，只运行一次

2 个答案: