mysql - 如何使用 INNER JOIN 进行选择，给出相关但不相关的记录

Question

我有这两张桌子：

用户

+----+-------+
| id | name  |
+----+-------+
| 1  | John  |
+----+-------+
| 2  | Peter |
+----+-------+
| 3  | Lucas |
+----+-------+

礼物

+----+--------+----------+----------+
| id | boy_id | type     | quantity |
+----+--------+----------+----------+
| 1  | 1      | clothing | 3        |
+----+--------+----------+----------+
| 2  | 2      | toy      | 1        |
+----+--------+----------+----------+
| 3  | 2      | clothing | 2        |
+----+--------+----------+----------+

我试图查询“男孩收到的礼物”，如下所示：

+-------+----------+----------+
| name  | type     | quantity |
+-------+----------+----------+
| John  | clothing | 3        |
+-------+----------+----------+
| Peter | toy      | 1        |
+-------+----------+----------+
| Peter | clothing | 2        |
+-------+----------+----------+
| Lucas | ""       | 0        |
+-------+----------+----------+

但我只知道如何通过这样的查询获取相关数据：

SELECT u.name, g.type, g.quantity FROM gifts g INNER JOIN users u ON g.boy_id = u.id

因此，由于卢卡斯不在礼品表中，我得到以下信息：

+-------+----------+----------+
| name  | type     | quantity |
+-------+----------+----------+
| John  | clothing | 3        |
+-------+----------+----------+
| Peter | toy      | 1        |
+-------+----------+----------+
| Peter | clothing | 2        |
+-------+----------+----------+

那么，如果男孩没有收到礼物，我如何在结果中包含彼得，用“”和零填写类型和数量？

标签： mysqlsqlleft-joinsql-null

Answer 1

你可以left join：

SELECT u.name, g.type, COALESCE(g.quantity, 0) quantity
FROM users u  
LEFT JOIN gifts g ON g.boy_id = u.id