sql语句左外连接和右外连接的区别

最新推荐文章于 2023-07-02 00:51:03 发布

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最新推荐文章于 2023-07-02 00:51:03 发布

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分类专栏：数据库文章标签： sql null mysql join c

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本文介绍了MySQL中的左外连接和右外连接操作，并通过具体例子展示了如何使用这些连接来查询两个表之间的相关数据。同时，还展示了如何进行分组汇总查询。

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mysql> select * from customers;
+----+-------+------+
| id | name | age |
+----+-------+------+
| 1 | tom   |   21 |
| 3 | Mike |   24 |
| 4 | Jack |   30 |
| 5 | Linda |   25 |
| 6 | Tom   | NULL |
+----+-------+------+

mysql> select * from orders;
+----+----------------+-------+-------------+
| id | order_number   | price | customer_id |
+----+----------------+-------+-------------+
| 1 | Tom_Order001   |   100 |           1 |
| 2 | Tom_Order002   |   200 |           1 |
| 3 | Tom_Order003   |   300 |           1 |
| 4 | Mike_Order001 |   100 |           3 |
| 5 | Jack_Order001 |   200 |           4 |
| 6 | Linda_Order001 |   100 |           5 |
| 7 | UnknownOrder   |   200 |        NULL |
+----+----------------+-------+-------------+

左外连接：
如果用左外连接的话，指查询出来的是在右表中没有对应左表中的记录以及符合条件（c.id=o.customer_id）的数据，比如：
  customers为左表，orders为右表
  select c.id,o.customer_id,c.name,o.id order_id,order_number from customers c left outer join orders o on c.id=o.customer_id;
  这样的结果显示的是在左表中的某些用户在orders表中没有对应的订单

mysql> select c.id,o.customer_id,c.name,o.id order_id,order_number from customers c left outer join orders o on c.id=o.customer_id;
+----+-------------+-------+----------+----------------+
| id | customer_id | name | order_id | order_number   |
+----+-------------+-------+----------+----------------+
| 1 |           1 | tom   |        1 | Tom_Order001   |
| 1 |           1 | tom   |        2 | Tom_Order002   |
| 1 |           1 | tom   |        3 | Tom_Order003   |
| 3 |           3 | Mike |        4 | Mike_Order001 |
| 4 |           4 | Jack |        5 | Jack_Order001 |
| 5 |           5 | Linda |        6 | Linda_Order001 |
| 6 |        NULL | Tom   |     NULL | NULL           |
+----+-------------+-------+----------+----------------+
7 rows in set (0.00 sec)

右外连接：
正好和左外连接相反，查询出来的结果是在左表中没有对应的项和以及符合条件（c.id=o.customer_id）的数据

mysql> select c.id,o.customer_id,c.name,o.id order_id,order_number from customers c right outer join orders o on c.id=o.customer_id;
+------+-------------+-------+----------+----------------+
| id   | customer_id | name | order_id | order_number   |
+------+-------------+-------+----------+----------------+
|    1 |           1 | tom   |        1 | Tom_Order001   |
|    1 |           1 | tom   |        2 | Tom_Order002   |
|    1 |           1 | tom   |        3 | Tom_Order003   |
|    3 |           3 | Mike |        4 | Mike_Order001 |
|    4 |           4 | Jack |        5 | Jack_Order001 |
|    5 |           5 | Linda |        6 | Linda_Order001 |
| NULL |        NULL | NULL |        7 | UnknownOrder   |
+------+-------------+-------+----------+----------------+
7 rows in set (0.00 sec)

=============================================================================

mysql> select name,age,sum(price) from customers c left join orders o on c.id=o.customer_id group by c.id;
+-------+------+------------+
| name | age | sum(price) |
+-------+------+------------+
| tom   |   21 |        600 |
| Mike |   24 |        100 |
| Jack |   30 |        200 |
| Linda |   25 |        100 |
| Tom   | NULL |       NULL |
+-------+------+------------+
5 rows in set (0.00 sec)