之所以要说这个问题,是因为项目中用到了not exists,但两者写的语句只有一点差别,结果一个有问题了,一个没问题。具体问题下面详细说明,先来看看exists如何应用。
exists:
强调的是是否有返回集,不需知道具体返回的是什么,比如:
SELECT * FROM customer WHERE not EXISTS ( SELECT 0 FROM customer_goods WHERE customer_id = 1 )
只要exists引导的子句有结果集返回,这个条件就算成立。这个返回的字段始终是0,改成1,则始终返回的是1,所以exists不
在乎返回的是什么内容,只在乎是否有结果集返回。
exists 和in 的区别
这二者最大的区别,是使用in只能返回一个字段值
SELECT * FROM customer c WHERE c.id not in ( SELECT customer_id FROM customer_goods WHERE customer_id = 1 )
但exists允许返回多个字段。
not in 和not exists 分别为in 和exists的对立面。
exists(sql 返回结果集为真)
not exists(sql 不返回结果集为真)
not exists详细介绍:
表customer:
表customer_goods:
二者的干系:customer_goods.customer_id = customer.id
(1) 查询:
SELECT * FROM customer c WHERE NOT EXISTS ( SELECT * FROM customer_goods cg WHERE cg.customer_id =1 )
结果:
无返回结果
(2)查询:
SELECT * FROM customer c WHERE NOT EXISTS ( SELECT * FROM customer_goods cg WHERE c.id =1 )
结果:
(3)分析:
发现二者差别只是是否not exists字句查询的查询条件是否跟外面查询条件有关,如果not exists子查询只有自己本身的查询条件,这样只要子查询中有数据返回,就证明是false,结果在整体执行就无返回值;一旦跟外面的查询关联上,就能准确查出数据。
而我遇到的问题正是这个。
经过分析,我认为一旦跟外层查询关联上,就会扫描外面查询的表。而没一旦二者不添加关联关系,只会根据not exists返回是否有结果集来判断,这也是为什么一旦子查询有数据,就查不到所有的数据了。
附案例分析
来看看not exists或exists是如何用的吧。
# 学生表 CREATE TABLE `Student`( `s_id` VARCHAR(20), `s_name` VARCHAR(20) NOT NULL DEFAULT "", `s_birth` VARCHAR(20) NOT NULL DEFAULT "", `s_sex` VARCHAR(10) NOT NULL DEFAULT "", PRIMARY KEY(`s_id`) ); # 课程表 CREATE TABLE `Course`( `c_id` VARCHAR(20), `c_name` VARCHAR(20) NOT NULL DEFAULT "", `t_id` VARCHAR(20) NOT NULL, PRIMARY KEY(`c_id`) ); # 教师表 CREATE TABLE `Teacher`( `t_id` VARCHAR(20), `t_name` VARCHAR(20) NOT NULL DEFAULT "", PRIMARY KEY(`t_id`) ); # 成绩表 CREATE TABLE `Score`( `s_id` VARCHAR(20), `c_id` VARCHAR(20), `s_score` INT(3), PRIMARY KEY(`s_id`,`c_id`) ); # 插入学生表测试数据 insert into Student values("01" , "赵雷" , "1990-01-01" , "男"); insert into Student values("02" , "钱电" , "1990-12-21" , "男"); insert into Student values("03" , "孙风" , "1990-05-20" , "男"); insert into Student values("04" , "李云" , "1990-08-06" , "男"); insert into Student values("05" , "周梅" , "1991-12-01" , "女"); insert into Student values("06" , "吴兰" , "1992-03-01" , "女"); insert into Student values("07" , "郑竹" , "1989-07-01" , "女"); insert into Student values("08" , "王菊" , "1990-01-20" , "女"); #课程表测试数据 insert into Course values("01" , "语文" , "02"); insert into Course values("02" , "数学" , "01"); insert into Course values("03" , "英语" , "03"); # 教师表测试数据 insert into Teacher values("01" , "张三"); insert into Teacher values("02" , "李四"); insert into Teacher values("03" , "王五"); #成绩表测试数据 insert into Score values("01" , "01" , 80); insert into Score values("01" , "02" , 90); insert into Score values("01" , "03" , 99); insert into Score values("02" , "01" , 70); insert into Score values("02" , "02" , 60); insert into Score values("02" , "03" , 80); insert into Score values("03" , "01" , 80); insert into Score values("03" , "02" , 80); insert into Score values("03" , "03" , 80); insert into Score values("04" , "01" , 50); insert into Score values("04" , "02" , 30); insert into Score values("04" , "03" , 20); insert into Score values("05" , "01" , 76); insert into Score values("05" , "02" , 87); insert into Score values("06" , "01" , 31); insert into Score values("06" , "03" , 34); insert into Score values("07" , "02" , 89); insert into Score values("07" , "03" , 98);
题目是查询和"01"号的同学学习的课程完全相同的其他同学的信息,直接做确实有点麻烦,我们可以先做做这题:查看学了所有课程的同学的信息。
学了所有课程的同学的信息,那不就是这些同学没有一门课程没有学吗。
select * from Student st where not exists(select * from Course c where not exists(select * from Score sc where sc.c_id = c.c_id and sc.s_id = st.s_id));
然后我们再回过来看这题,把所有的课程换成01同学学的课程。
select * from Student st where not exists(select * from ( select s2.c_id as c_id from Student s1 inner join Score s2 on s1.s_id = s2.s_id where s1.s_id = 01) t where not exists (select * from Score sc where sc.c_id = t.c_id and sc.s_id = st.s_id and st.s_id != 01));
总结
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原文链接:https://blog.csdn.net/zhangsify/article/details/71937745