Question

我有两张桌子

card_index

+-----+-----+-----+-----+
| uid | id1 | id2 | id3 |
+-----+-----+-----+-----+
|  1  | 125 | 129 | 145 |
|  2  | 127 | 128 | 132 |
+-----+-----+-----+-----+

和card_data

+----+------------+----------+...
| id |  last_use  |  active  |
+----+------------+----------+...
| 125| 10-07-2014 |   true   |
| 127| 27-02-2014 |   true   |
| 128| 15-06-2014 |   true   |
| 129| 10-01-2013 |   false  |
+----+------------+----------+... and so on

我想从'card_index'中选择'card_data'中'uid'的所有数据，因此'uid'= 1的结果应如下所示：

+----+------------+----------+
| id |  last_use  |  active  |
+----+------------+----------+
| 125| 10-07-2014 |   true   |
| 129| 10-01-2013 |   false  |
| 145| 13-01-2013 |   true   |
+----+------------+----------+

我做的SQL查询是：

SELECT * FROM card_data WHERE id IN (SELECT CONCAT(id1,',',id2,',',id3) FROM card_index WHERE uid = 1);

然而，它只选择第一行：

+----+------------+----------+
| id |  last_use  |  active  |
+----+------------+----------+
| 125| 10-07-2014 |   true   |
+----+------------+----------+

任何人都可以帮助我吗？我不知道出了什么问题。我正在使用从FreeBSD上的源代码编译的mysql-server 5.5.38-log

Answer 1

使用INNER JOIN代替CONCAT和WHERE id IN的组合为我工作。这是查询：

SELECT * FROM card_data
INNER JOIN card_index ON id1=id OR id2=id OR id3=id
WHERE uid = 1;

期望的结果：

+----+------------+----------+
| id |  last_use  |  active  |
+----+------------+----------+
| 125| 10-07-2014 |   true   |
| 129| 10-01-2013 |   false  |
| 145| 13-01-2013 |   true   |
+----+------------+----------+

MySQL选择多个列为一列

1 个答案: