本篇内容介绍了“MySQL的leftjoin、rightjoin、innerjoin的案例分析”的有关知识,在实际案例的操作过程中,不少人都会遇到这样的困境,接下来就让小编带领大家学习一下如何处理这些情况吧!希望大家仔细阅读,能够学有所成!
left join(左联接) 返回包括左表中的所有记录和右表中联结字段相等的记录
right join(右联接) 返回包括右表中的所有记录和左表中联结字段相等的记录
inner join(等值连接) 只返回两个表中联结字段相等的行
举例如下:
CREATE TABLE a(
aID int,
aNum char(20)
) ;
CREATE TABLE b(
bID int,
bName char(20)
) ;
INSERT INTO a
VALUES ( 1, 'a20050111' ) , ( 2, 'a20050112' ) , ( 3, 'a20050113' ) , ( 4, 'a20050114' ) , ( 5, 'a20050115' ) ;
INSERT INTO b
VALUES ( 1, '2006032401' ) , ( 2, '2006032402' ) , ( 3, '2006032403' ) , ( 4, '2006032404' ) , ( 8, '2006032408' ) ;
表A记录如下:
aID aNum
1 a20050111
2 a20050112
3 a20050113
4 a20050114
5 a20050115
表B记录如下:
bID bName
1 2006032401
2 2006032402
3 2006032403
4 2006032404
8 2006032408
--------------------------------------------
1.left join
mysql> select * from A
-> left join B
-> on A.aID = B.bID;
+------+-----------+------+------------+
| aID | aNum | bID | bName |
+------+-----------+------+------------+
| 1 | a20050111 | 1 | 2006032401 |
| 2 | a20050112 | 2 | 2006032402 |
| 3 | a20050113 | 3 | 2006032403 |
| 4 | a20050114 | 4 | 2006032404 |
| 5 | a20050115 | NULL | NULL |
+------+-----------+------+------------+
(所影响的行数为 5 行)
结果说明:
left join是以A表的记录为基础的,A可以看成左表,B可以看成右表,left join是以左表为准的.
换句话说,左表(A)的记录将会全部表示出来,而右表(B)只会显示符合搜索条件的记录(例子中为: A.aID = B.bID).
B表记录不足的地方均为NULL.
--------------------------------------------
2.right join
mysql> select * from A
-> right join B
-> on A.aID = B.bID;
+------+-----------+------+------------+
| aID | aNum | bID | bName |
+------+-----------+------+------------+
| 1 | a20050111 | 1 | 2006032401 |
| 2 | a20050112 | 2 | 2006032402 |
| 3 | a20050113 | 3 | 2006032403 |
| 4 | a20050114 | 4 | 2006032404 |
| NULL | NULL | 8 | 2006032408 |
+------+-----------+------+------------+
5 rows in set (0.01 sec)
(所影响的行数为 5 行)
结果说明:
仔细观察一下,就会发现,和left join的结果刚好相反,这次是以右表(B)为基础的,A表不足的地方用NULL填充.
--------------------------------------------
3.inner join
mysql> select * from A inner join B on A.aID = B.bID;
+------+-----------+------+------------+
| aID | aNum | bID | bName |
+------+-----------+------+------------+
| 1 | a20050111 | 1 | 2006032401 |
| 2 | a20050112 | 2 | 2006032402 |
| 3 | a20050113 | 3 | 2006032403 |
| 4 | a20050114 | 4 | 2006032404 |
+------+-----------+------+------------+
4 rows in set (0.00 sec)
结果说明:
很明显,这里只显示出了 A.aID = B.bID的记录.这说明inner join并不以谁为基础,它只显示符合条件的记录.
--------------------------------------------
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