Oracle exists/in和not exists/not in之前的区别与联系
曾写过一篇关于Oracle NULL对in和not in结果的影响：Oracle的where条件in/not in中包含NULL时的处理。今天来看看exists和not exists中NULL值对结果的影响。
网上经常看到关于in和exixts、not in和not exists性能比对和互换的例子，但它们真的就可以简单互换么？我们通过下面的实验来看一下。
实验环境：Oracle 11.2.0.4
1、创建表并插入测试数据
create table t1 (id number);
create table t2 (id number);
insert into t1 values(1);
insert into t1 values(2);
insert into t1 values(3);
insert into t1 values(4);
insert into t1 values(null);
commit;
insert into t2 values(3);
insert into t2 values(4);
insert into t2 values(5);
insert into t2 values(6);
commit;
linuxidc@ORA11G>select * from t1;
ID
----------
1
2
3
4
5 rows selected.
linuxidc@ORA11G>select * from t2;
ID
----------
3
4
5
6
4 rows selected.
第一种情况：exists/in的查询中不包含NULL，外层查询包含NULL
linuxidc@ORA11G>select * from t1 where exists(select 1 from t2 where t1.id=t2.id);
ID
----------
3
4
2 rows selected.
linuxidc@ORA11G>select * from t1 where id in (select id from t2);
ID
----------
3
4
2 rows selected.
从上面的查询结果看出exists和in都查到了id=2和3的两条数据。
第二种情况：not exists/not in的查询中不包含NULL，外层查询包含NULL
linuxidc@ORA11G>select * from t1 where not exists(select 1 from t2 where t1.id=t2.id);
ID
----------
1
2
3 rows selected.
linuxidc@ORA11G>select * from t1 where id not in (select id from t2);
ID
----------
1
2
2 rows selected.
从上面的结果中可以看到两个查询都查到了id=1和2这两条记录，但not exists还查到了t1表中id为NULL的行。原因是表t1中id为NULL的行exists(3,4,5,6)为False，但前面加了个not则返回结果就为True了。
第三种情况：exists/in的子查询中包含NULL，外层查询包含NULL
linuxidc@ORA11G>insert into t2 values(null);
1 row created.
linuxidc@ORA11G>commit;
Commit complete.
linuxidc@ORA11G>select * from t1 where id in (select id from t2);
ID
----------
3
4
2 rows selected.
linuxidc@ORA11G>select * from t1 where exists(select 1 from t2 where t1.id=t2.id);
ID
----------
3
4
2 rows selected.
从上面的结果中可以看出exist和in的结果是一致的。
第四种情况：not exists和not in的查询中包含NULL
linuxidc@ORA11G>select * from t1 where not exists(select 1 from t2 where t1.id=t2.id);
ID
----------
1
2
3 rows selected.
linuxidc@ORA11G>select * from t1 where id not in (select id from t2);
no rows selected
从上面的查询结果中可以看出两个结果差异很大，not exists把id=1和2和为NULL的值都查出来了，而not in查出来的结果为空。no in结果为空的原因可以参考之前的文章，not exists的原因与第二种情况类似。
第五种情况：not in/not exists的子查询中无NULL值，外层查询也无NULL值
linuxidc@ORA11G>delete from t1 where id is null;
1 row deleted.
linuxidc@ORA11G>delete from t2 where id is null;
1 row deleted.
linuxidc@ORA11G>commit;
Commit complete.
linuxidc@ORA11G>select * from t1 where id not in (select id from t2);
ID
----------
1
2
2 rows selected.
linuxidc@ORA11G>select * from t1 where not exists(select 1 from t2 where t1.id=t2.id);
ID
----------
1
2
2 rows selected.
第六种情况：in/exists的子查询中无NULL值，外层查询也无NULL值
linuxidc@ORA11G>select * from t1 where id in (select id from t2);
ID
----------
3
4
2 rows selected.
linuxidc@ORA11G>select * from t1 where exists(select 1 from t2 where t1.id=t2.id);
ID
----------
3
4
2 rows selected.
第七种情况：in/exists的子查询中有NULL值，外层查询无NULL值
linuxidc@ORA11G>insert into t2 values(null);
1 row created.
linuxidc@ORA11G>commit;
Commit complete.
linuxidc@ORA11G>select * from t1 where id in (select id from t2);
ID
----------
3
4
2 rows selected.
linuxidc@ORA11G>select * from t1 where exists(select 1 from t2 where t1.id=t2.id);
ID
----------
3
4
2 rows selected.
第八种情况：not in/not exists的子查询中有NULL值，外层查询无NULL值
linuxidc@ORA11G>select * from t1 where id not in (select id from t2);
no rows selected
linuxidc@ORA11G>select * from t1 where not exists(select 1 from t2 where t1.id=t2.id);
ID
----------
1
2
2 rows selected.
从上面的八种情况我们可以总结如下：
1、in和exists在有无NULL的情况下可以相互转换。
2、not in和not exists在都没有NULL值的情况下才可以相互转换。