MySQL多对多选择
仍在学习MySQL的绳索,我试图找出如何进行涉及多对多的特定选择。如果表名太通用,我深表歉意,我只是在做一些自制的练习。我尽力成为一名自学者。
我有3个表,其中一个是链接表。如何编写
(他们拥有2部手机)的语句。我猜答案在WHERE语句中,但我不知道该怎么写。
-- Table: mark3+---------+-----------+
| phoneid | name |
+---------+-----------+
| 1 | HTC |
| 2 | Nokia |
| 3 | Samsung |
| 4 | Motorolla |
+---------+-----------+
-- Table: mark4
+------+---------+
| uid | phoneid |
+------+---------+
| 1 | 1 |
| 1 | 2 |
| 2 | 1 |
| 2 | 3 |
| 2 | 4 |
| 3 | 1 |
| 3 | 3 |
+------+---------+
-- Table: mark5
+------+-------+
| uid | name |
+------+-------+
| 1 | John |
| 2 | Paul |
| 3 | Peter |
+------+-------+
回答:
关键是在GROUP BY / HAVING中使用COUNT个DISTINCT电话名称。当计数为2时,您将知道用户同时拥有 电话。
SELECT m5.name FROM mark5 m5
INNER JOIN mark4 m4
ON m5.uid = m4.uid
INNER JOIN mark3 m3
ON m4.phoneid = m3.phoneid
WHERE m3.name in ('HTC', 'Samsung')
GROUP BY m5.name
HAVING COUNT(DISTINCT m3.name) = 2;
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