in和exists执行时,in是先执行子查询中的查询,然后再执行主查询。而exists查询它是先执行主查询,即外层表的查询,然后再执行子查询。
exists 和 in 在执行时效率单从执行时间来说差不多,exists要稍微优于in。在使用时一般应该是用exists而不用in
如果子查询得出的结果集记录较少,主查询中的表较大且又有索引时应该用in,反之如果外层的主查询记录较少,子查询中的表大,又有索引时使用exists。IN时不对NULL进行处理。
not exists 和 not in 比较时,not exists 的效率比较高。
为了说明测试结果,我把emp1表中的数据到了315392条。emp2中删除只有2条件数据。测试的依据是执行的时间来说明的。
emp1中的数据记录情况。
SQL> select count(*) from emp1;
COUNT(*)
----------
315392
emp2中的数据记录情况:
SQL> select count(*) from emp2;
COUNT(*)
----------
2
1、 执行exists查询,要求在emp1中查询出所有存在于emp2的数据总数
SQL> select count(*) from emp1 where exists ( select null from emp2 where emp1.ename = emp2.ename);
COUNT(*)
----------
45056
执行次数十次,最大的一次为0.125S
2、 使用not exists查询出所在不在emp2中的数据总数
SQL> select count(*) from emp1 where not exists ( select null from emp2 where emp1.ename = emp2.ename);
COUNT(*)
----------
270336
执行次数十次,最大的一次为0.141S
3、执行in 查询,要求在emp1中查询出所有存在于emp2的数据总数
SQL> select count(*) from emp1 where ename in ( select ename from emp2);
COUNT(*)
----------
45056
执行十次,最大的一次为0.141S
4、使用not in查询出所在不在emp2中的数据总数
SQL> select count(*) from emp1 where ename not in ( select ename from emp2 );
COUNT(*)
----------
270336
执行十次,最长一次为0.328S
5、使用in查询,调用外层与子查询的位置,要求查询出存在于emp2中,且存在于emp1中的数据记录数
SQL> select count(*) from emp2 where ename in (select ename from emp1 );
COUNT(*)
----------
2
执行次数十次,最长的一次为0.047S
6、使用exists查询,调用外层与子查询的位置,要求查询出存在于emp2中,且存在于emp1中的数据记录数
SQL> select count(*) from emp2 where ename in (select ename from emp1 );
COUNT(*)
----------
2
执行次数十次,最长的一次为0.047S
综上所述:在使用in 和 exists时,个人觉得,效率差不多。而在not in 和 not exists比较时,not exists的效率要比not in的效率要高。
当使用in时,子查询where条件不受外层的影响,自动优化会转成exist语句,它的效率和exist一样。(没有验证)
如select * from t1 where f1 in (select f1 from t2 where t2.fx='x') 这时,认为in 和 exists效率一样。
IN适合于外表大而内表小的情况;EXISTS适合于外表小而内表大的情况。

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