SELECT ID,NAME FROM A WHERE EXIST (SELECT * FROM B WHERE A.ID=B.AID)
执行结果为
1 A1
2 A2
原因可以按照如下分析
SELECT ID,NAME FROM A WHERE EXISTS (SELECT * FROM B WHERE B.AID=１)
--->SELECT * FROM B WHERE B.AID=１有值返回真所以有数据
SELECT ID,NAME FROM A WHERE EXISTS (SELECT * FROM B WHERE B.AID=2)
--->SELECT * FROM B WHERE B.AID=２有值返回真所以有数据
SELECT ID,NAME FROM A WHERE EXISTS (SELECT * FROM B WHERE B.AID=3)
--->SELECT * FROM B WHERE B.AID=３无值返回真所以没有数据
NOT EXISTS 就是反过来
SELECT ID,NAME FROM A WHERE NOT EXIST (SELECT * FROM B WHERE A.ID=B.AID)
执行结果为
3 A3
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EXISTS = IN,意思相同不过语法上有点点区别，好像使用IN效率要差点，应该是不会执行索引的原因
SELECT ID,NAME FROM A WHERE ID IN (SELECT AID FROM B)
NOT EXISTS = NOT IN ,意思相同不过语法上有点点区别
SELECT ID,NAME FROM A WHERE ID NOT IN (SELECT AID FROM B)
