转载自:/johnsmith/archive//09/07/2169407.html
参考:/view/577f4d49cf84b9d528ea7a6f.html //这个讲的很详细
引用自:http://chenling1018./blog/static/1480254112944944177/
--exists (sql 返回结果集,为真) --not exists (sql 不返回结果集,为真) --如下:
--表A ID NAME1A12A23A3
-- 表B
ID AID NAME
11B1
22B2
32B3
-- 表A和表B是1对多的关系 A.ID => B.AID
SELECT ID,NAME FROM A WHERE EXIST ( SELECT * FROM B WHERE A.ID = B.AID)
-- 执行结果为
1A1
2A2
-- 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,返回假,所以没有数据
--NOT EXISTS 就是反过来 SELECT ID,NAME FROM A WHERENOT EXIST (SELECT * FROM B WHERE A.ID=B.AID) --执行结果为3A3
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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 WHEREID 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 WHEREIDNOT IN (SELECT AID FROM B)