SQL中left join后面on、where的区别
创建两张表并插入一些数据
create table class( class_id int, class_name varchar(20), class_grade char(1) ); insert into class values (1,'语文','A'); insert into class values (2,'数学','B'); insert into class values (3,'英语','C'); create table score( class_id int, stu_id varchar(20), Score int ); insert into score values (1,'A001',91); insert into score values (2,'A001',95); insert into score values (1,'A002',82); insert into score values (2,'A002',87); insert into score values (3,'B003',65);
比较下面几组查询结果:
select * from class A left join score B on A.class_id=B.class_id; +----------+------------+-------------+----------+--------+-------+ | class_id | class_name | class_grade | class_id | stu_id | Score | +----------+------------+-------------+----------+--------+-------+ | 1 | 语文 | A | 1 | A001 | 91 | | 2 | 数学 | B | 2 | A001 | 95 | | 1 | 语文 | A | 1 | A002 | 82 | | 2 | 数学 | B | 2 | A002 | 87 | | 3 | 英语 | C | 3 | B003 | 65 | +----------+------------+-------------+----------+--------+-------+ 5 rows in set (0.00 sec)
select * from class A left join score B on 1=0; +----------+------------+-------------+----------+--------+-------+ | class_id | class_name | class_grade | class_id | stu_id | Score | +----------+------------+-------------+----------+--------+-------+ | 1 | 语文 | A | NULL | NULL | NULL | | 2 | 数学 | B | NULL | NULL | NULL | | 3 | 英语 | C | NULL | NULL | NULL | +----------+------------+-------------+----------+--------+-------+ 3 rows in set (0.00 sec)
left join的最重要特点是:不管on后面是什么条件,都会返回左表中的所有行!
select * from class A left join score B on A.class_id=B.class_id and A.class_name='语文'; +----------+------------+-------------+----------+--------+-------+ | class_id | class_name | class_grade | class_id | stu_id | Score | +----------+------------+-------------+----------+--------+-------+ | 1 | 语文 | A | 1 | A001 | 91 | | 1 | 语文 | A | 1 | A002 | 82 | | 2 | 数学 | B | NULL | NULL | NULL | | 3 | 英语 | C | NULL | NULL | NULL | +----------+------------+-------------+----------+--------+-------+ 4 rows in set (0.00 sec)
select * from class A left join score B on A.class_id=B.class_id and A.class_name='数学'; +----------+------------+-------------+----------+--------+-------+ | class_id | class_name | class_grade | class_id | stu_id | Score | +----------+------------+-------------+----------+--------+-------+ | 2 | 数学 | B | 2 | A001 | 95 | | 2 | 数学 | B | 2 | A002 | 87 | | 1 | 语文 | A | NULL | NULL | NULL | | 3 | 英语 | C | NULL | NULL | NULL | +----------+------------+-------------+----------+--------+-------+ 4 rows in set (0.00 sec)
select * from class A left join score B on A.class_id=B.class_id and A.class_name='体育'; +----------+------------+-------------+----------+--------+-------+ | class_id | class_name | class_grade | class_id | stu_id | Score | +----------+------------+-------------+----------+--------+-------+ | 1 | 语文 | A | NULL | NULL | NULL | | 2 | 数学 | B | NULL | NULL | NULL | | 3 | 英语 | C | NULL | NULL | NULL | +----------+------------+-------------+----------+--------+-------+ 3 rows in set (0.00 sec)
如果on后面的条件是左表中的列,那么左表中满足条件的行和右表中的行进行匹配,左表中不满足条件的行其右表中为Null。
select * from class A left join score B on A.class_id=B.class_id and B.Score=90; +----------+------------+-------------+----------+--------+-------+ | class_id | class_name | class_grade | class_id | stu_id | Score | +----------+------------+-------------+----------+--------+-------+ | 1 | 语文 | A | NULL | NULL | NULL | | 2 | 数学 | B | NULL | NULL | NULL | | 3 | 英语 | C | NULL | NULL | NULL | +----------+------------+-------------+----------+--------+-------+ 3 rows in set (0.01 sec)
select * from class A left join score B on A.class_id=B.class_id and B.Score=65; +----------+------------+-------------+----------+--------+-------+ | class_id | class_name | class_grade | class_id | stu_id | Score | +----------+------------+-------------+----------+--------+-------+ | 3 | 英语 | C | 3 | B003 | 65 | | 1 | 语文 | A | NULL | NULL | NULL | | 2 | 数学 | B | NULL | NULL | NULL | +----------+------------+-------------+----------+--------+-------+ 3 rows in set (0.01 sec)
如果on后面的条件是右表中的列(and rightTable.colName='***'),首先会根据(and rightTable.colName='***')过滤掉右表中不满足条件的行;然后,左表中的行根据(on leftTable.id=rightTable.id)和右表中满足条件的行进行匹配。
select * from class A left join score B on A.class_id=B.class_id and A.class_name='体育' and B.Score=82; +----------+------------+-------------+----------+--------+-------+ | class_id | class_name | class_grade | class_id | stu_id | Score | +----------+------------+-------------+----------+--------+-------+ | 1 | 语文 | A | NULL | NULL | NULL | | 2 | 数学 | B | NULL | NULL | NULL | | 3 | 英语 | C | NULL | NULL | NULL | +----------+------------+-------------+----------+--------+-------+ 3 rows in set (0.00 sec)
/**********************过滤条件在on中时**********************/
总结一下,如果 left join on leftTable.id=rightTable.id 后还有其他条件:
(1)and leftTable.colName='***',过滤左表,但是左表不满足条件的行直接输出,并将右表对应部分置为null
(2)and rightTable.colName='***',过滤右表,对左表没有影响
(3)and leftTable.colName='***' and rightTable.colName='***',就是上面(1)和(2)一起发挥作用
不管on后面有哪些条件,left join都要返回左表中的所有行!
/**********************过滤条件在where中时**********************/
过滤条件写在where中时,先执行left join,然后再根据where条件对表进行过滤
posted on 2019-03-28 20:14 Muliti_Hu 阅读(1122) 评论(0) 编辑 收藏 举报