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。如何撰写声明“显示哪些用户同时拥有HTC和三星手机”（他们拥有2部手机）。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

密钥在GROUP BY / HAVING中使用COUNT个DISTINCT电话名称。当计数为2时，您将知道用户两个电话。

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;

Answer 2

我认为您正在寻找的答案是：

"SELECT DISTINCT mark5.name FROM mark5 JOIN mark4 ON (mark4.uid = mark5.uid) JOIN mark3 ON (mark3.phoneid = mark4.phoneid) WHERE mark3.name = 'HTC' && mark3.name = 'Samsung'"

我相信，如果您从mark5中提取不同的名称，并在电话名称为HTC或三星的特定值上加入其他表格，我会回答您的问题

MySQL多对多选择

2 个答案: