MySQL多对多选择

Question

仍在学习MySQL绳索,我正试图找出如何进行涉及多对多的特定选择.如果表名太通用我很抱歉,我只是做了一些自制的练习.我尽我所能成为一名自学者.

我有3个表,其中一个是链接表.如何编写"显示哪些用户同时拥有HTC和三星手机"(他们拥有2部手机)的声明.我猜测答案是在WHERE语句中,但我无法弄清楚如何说出来.

-- 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 |
+------+-------+

Answer 1

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两部手机.

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;