将SQL表头值与另一个表列的值进行匹配

Question

下面有两个数据库表Table_1 & Table_2。Table_1列索引与table_2的id值匹配。Table_2的Name值将作为查询结果的列标题。

Table_1
______________________________________________
| date      |city_1 | city_2 | ... | city_100 |
|-----------|-------|-------------------------|
| 20.02.2013|   4   |   34   | ... |   222    |
| 21.02.2013|   3   |   10   | ... |    33    |
|    ...    |  ...  |   ...  | ... |   ...    |
|_____________________________________________|


  Table_2
___________________
|  id   |   name  |
|-------|---------|
|   1   | newyork |
|   2   | london  |
|  ...  |   ...   |
|  100  | istanbul|
|_________________|

预期结果如下

  __________________________________________________________
  |  date     |  newyork   |   london   |  ...  |  istanbul |
  |-----------|------------|------------|-------|-----------|
  | 20.02.2013|     4      |     34     |  ...  |    222    |
  | 21.02.2013|     3      |     10     |  ...  |     33    |
  |   ...     |    ...     |    ...     |  ...  |    ...    |
  |___________|____________|____________|_______|___________| 

要获得上述结果的SQL查询是什么？

谢谢

Answer 1

您可以使用这样的解决方案：

SET @sql = NULL;

SELECT GROUP_CONCAT(COALESCE(CONCAT(COLUMN_NAME, ' as ', Table_2.Name), COLUMN_NAME))
FROM
  INFORMATION_SCHEMA.COLUMNS
  LEFT JOIN Table_2
  ON Table_2.ID = SUBSTRING_INDEX(COLUMN_NAME, '_', -1)
WHERE table_name = 'Table_1' INTO @sql;

SET @sql = CONCAT('SELECT ', @sql, ' FROM Table_1');

PREPARE stmt FROM @sql;
EXECUTE stmt;
DEALLOCATE PREPARE stmt;

看看吧，here。这段代码将提取Table_1的所有列名，并尝试将列名与Table_2中的in连接起来。如果匹配，我将返回带有别名的列名，如下所示：

city_1 AS newyour

使用GROUP_CONCAT，变量@sql将包含所有列，如下所示：

date,city_1 as newyork,city_2 as london,city_100 as istanbul

然后，我使用别名+ ' FROM Table_1'连接'SELECT ' +列名，并执行结果查询。