将SQL表头值与另一个表列值匹配

时间:2013-02-20 10:37:32

标签: mysql sql

下面有2个数据库表,Table_1& TABLE_2。 Table_1列索引与table_2的id值匹配。 Table_2的名称值将是查询结果的列标题。

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|
|_________________|

预期结果低于

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

获取上述结果的SQL查询是什么?

由于

1 个答案:

答案 0 :(得分:2)

你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;

here。 此代码将提取Table_1的所有列名称,并尝试使用Table_2中的ID连接列的名称。如果匹配,我将使用别名返回列的名称,如下所示:

city_1 AS newyour

并使用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

然后我将'SELECT ' +列名与别名+ ' FROM Table_1'连接起来,并执行生成的查询。