MySQL多对多选择 [英] MySQL Many-To-Many Select

问题描述

仍在学习MySQL的绳索，我试图找出如何进行涉及多对多选择的特定选择.如果表名太通用，我深表歉意，我只是在做一些自制的练习.我会尽力成为一名自学者.

Still learning the MySQL ropes and I'm trying to find out how to do a specific selection involving many-to-many. I apologize if the table names are too generic, I was just doing some self-made exercises. I try my best to be a self-learner.

我有3个表，其中一个是链接表.如何写显示哪些用户同时拥有HTC和Samsung手机" (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.

I have 3 tables one of which is a linking table. How do I write the statement which says "Show which users own both an HTC and a Samsung phone" (they own 2 phones). I'm guessing the answer is in the WHERE statement but I can't figure out how to word it.

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

推荐答案

密钥在GROUP BY/HAVING中，使用COUNT个DISTINCT电话名.当计数为2时，您将知道该用户同时拥有两个手机.

The key is in the GROUP BY/HAVING using a COUNT of DISTINCT phone names. When the count is 2, you'll know the user has both phones.

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;

这篇关于MySQL多对多选择的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！