Question

mysql> select * from product;
+------------+---------------+
| product_id | name          |
+------------+---------------+
|          1 | Car           | 
|          2 | House         | 
|          3 | Cat           | 
|          4 | Blank Product | 
+------------+---------------+
4 rows in set (0.00 sec)

mysql> select * from tag;
+--------+-----------+
| tag_id | name      |
+--------+-----------+
|      1 | Expensive | 
|      2 | Fast      | 
|      3 | Mean      | 
|      4 | Large     | 
|      5 | Small     | 
|      6 | Alive     | 
|      7 | Blank Tag | 
+--------+-----------+
7 rows in set (0.00 sec)

mysql> select * from product_tag;
+------------+--------+
| product_id | tag_id |
+------------+--------+
|          1 |      1 | 
|          1 |      2 | 
|          1 |      3 | 
|          1 |      4 | 
|          2 |      1 | 
|          2 |      4 | 
|          3 |      2 | 
|          3 |      3 | 
|          3 |      5 | 
|          3 |      6 | 
+------------+--------+
10 rows in set (0.00 sec)

为什么以下查询会返回我的空白标记但不返回空白产品？

mysql> select * from product_tag right join product using (product_id)
                                 right join tag using (tag_id);
+--------+-----------+------------+-------+
| tag_id | name      | product_id | name  |
+--------+-----------+------------+-------+
|      1 | Expensive |          1 | Car   | 
|      1 | Expensive |          2 | House | 
|      2 | Fast      |          1 | Car   | 
|      2 | Fast      |          3 | Cat   | 
|      3 | Mean      |          1 | Car   | 
|      3 | Mean      |          3 | Cat   | 
|      4 | Large     |          1 | Car   | 
|      4 | Large     |          2 | House | 
|      5 | Small     |          3 | Cat   | 
|      6 | Alive     |          3 | Cat   | 
|      7 | Blank Tag |       NULL | NULL  | 
+--------+-----------+------------+-------+
11 rows in set (0.00 sec)

Answer 1

您正在使用正确的加入。在您的查询标签中，ids是MySQL开始匹配的基础。右连接从右到左进行评估。如果您将查询分为两部分。第一个是：

select * from product_tag right join tag using (tag_id);
+--------+-----------+------------+
| tag_id | name      | product_id |
+--------+-----------+------------+
|      1 | expensive |          1 | 
|      1 | expensive |          2 | 
|      2 | fast      |          1 | 
|      2 | fast      |          3 | 
|      3 | mean      |          1 | 
|      3 | mean      |          3 | 
|      4 | larg      |          1 | 
|      4 | larg      |          2 | 
|      5 | small     |          3 | 
|      6 | alive     |          3 | 
|      7 | blank tag |       NULL | 

+--------+-----------+------------+

如您所见，没有与空白标签匹配的product_id。解释为什么将此结果与产品表结合将为您提供所见的结果。

如果您使用左连接，则会得到以下结果：

select * from product_tag left join product using (product_id) left join tag using (tag_id);

+--------+------------+-------+-----------+
| tag_id | product_id | name  | name      |
+--------+------------+-------+-----------+
|      1 |          1 | car   | expensive | 
|      2 |          1 | car   | fast      | 
|      3 |          1 | car   | mean      | 
|      4 |          1 | car   | larg      | 
|      1 |          2 | house | expensive | 
|      4 |          2 | house | larg      | 
|      2 |          3 | cat   | fast      | 
|      3 |          3 | cat   | mean      | 
|      5 |          3 | cat   | small     | 
|      6 |          3 | cat   | alive     | 
+--------+------------+-------+-----------+

Answer 2

没有将产品ID 4与标签相关联的行。您需要在product_tag表中添加一行，如下所示：

+------------+--------+
| product_id | tag_id |
+------------+--------+
|          4 |      7 | 
+------------+--------+

Answer 3

您正在使用RIGHT JOIN，因此即使连接表中没有匹配项，也会返回右表中的所有行，即“tag”。

MySQL多个右连接

3 个答案: