Question

我是SQL的新手并遇到以下挑战，我左边有一个表，它将产品分配给一家公司并用ref_id表示（它来自另一个表，我为了提问而对其进行了简化））我有一个主列表。我想将公司“A”的所有产品与主列表进行比较，并显示公司A尚未订阅的所有产品。我正在使用JOIN并最终获得重复的列名称，这会引发我的PHP程序。以下是我为UNION，RIGHT JOIN，LEFT JOIN和NOT运算符尝试了几种语法，以下是我得到的最接近的，但是我得到了重复的列名，我只需要最后两列的最后一个表。请指教。

mysql> select ref_id, product_id, short_name from ap_company_product order by ref_id, product_id;
+--------+------------+------------+
| ref_id | product_id | short_name |
+--------+------------+------------+
|      2 | 10         | product 10 |
|      2 | 11         | product 11 |
|      2 | 12         | product 12 |
|      2 | 15         | product 15 |
|      2 | 17         | product 17 |
|      2 | 21         | product 21 |
|      3 | 11         | product 11 |
|      3 | 13         | product 13 |
|      3 | 17         | product 17 |
|      3 | 20         | product 20 |
+--------+------------+------------+
10 rows in set (0.00 sec)

THE MAIN LIST

mysql> select  product_id, short_name from ap_company_product_list;
+------------+-------------+
| product_id | short_name  |
+------------+-------------+
| 10         | product 10  |
| 11         | product 11  |
| 12         | product 12  |
| 13         | product 13  |
| 14         | product 14  |
| 15         | product 15  |
| 16         | product 16  |
| 17         | product 17  |
| 18         | product 18  |
| 19         | product 19  |
| 20         | product 20  |
| 21         | product 21  |
| 22         | product 22  |
+------------+-------------+
13 rows in set (0.00 sec)


Another SQL which is the actual results I need

SELECT  aa.product_id as product_id, aa.short_name as short_name, aa.ref_id as ref_id, bb.product_id as product_id, bb.short_name as short_name FROM (select ref_id, product_id, short_name from ap_company_product where ref_id=2) aa  RIGHT OUTER JOIN ap_company_product_list bb ON aa.product_id = bb.product_id where aa.product_id is NULL;
+------------+------------+--------+------------+------------+
| product_id | short_name | ref_id | product_id | short_name |
+------------+------------+--------+------------+------------+
| NULL       | NULL       |   NULL | 13         | product 13 |
| NULL       | NULL       |   NULL | 14         | product 14 |
| NULL       | NULL       |   NULL | 16         | product 16 |
| NULL       | NULL       |   NULL | 18         | product 18 |
| NULL       | NULL       |   NULL | 19         | product 19 |
| NULL       | NULL       |   NULL | 20         | product 20 |
| NULL       | NULL       |   NULL | 22         | product 22 |
+------------+------------+--------+------------+------------+
7 rows in set (0.01 sec)

Answer 1

请使用：

SELECT  bb.product_id as product_id, bb.short_name as short_name FROM (select ref_id, product_id, short_name from ap_company_product where ref_id=2) aa  RIGHT OUTER JOIN ap_company_product_list bb ON aa.product_id = bb.product_id where aa.product_id is NULL;

顺便说一下，我更喜欢这个：

SELECT product_id, short_name FROM ap_company_product_list WHERE product_id NOT IN (SELECT product_id FROM ap_company_product);

MYSQL Join返回重复的列名

1 个答案: