问题描述

我正在尝试获取一个原始数据集，为新数据添加列并将其转换为更传统的表结构.这个想法是让脚本提取列名称(日期)并将其放入一个新列中，然后将每个日期数据值堆叠在一起.

I'm trying to take a raw data set that adds columns for new data and convert it to a more traditional table structure. The idea is to have the script pull the column name (the date) and put that into a new column and then stack each dates data values on top of each other.

示例

Store     1/1/2013     2/1/2013
XYZ INC   $1000        $2000

到

Store     Date         Value
XYZ INC   1/1/2013     $1000
XYZ INC   2/1/2013     $2000

谢谢

推荐答案

有几种不同的方法可以获得您想要的结果.

There are a few different ways that you can get the result that you want.

您可以将 SELECT 与 UNION ALL 一起使用:

You can use a SELECT with UNION ALL:

select store, '1/1/2013' date, [1/1/2013] value
from yourtable
union all
select store, '2/1/2013' date, [2/1/2013] value
from yourtable;

参见SQL Fiddle with Demo.

您可以使用 UNPIVOT 函数:

You can use the UNPIVOT function:

select store, date, value
from yourtable
unpivot
(
  value
  for date in ([1/1/2013], [2/1/2013])
) un;

参见 SQL Fiddle with Demo.

最后，根据您的 SQL Server 版本，您可以使用 CROSS APPLY:

Finally, depending on your version of SQL Server you can use CROSS APPLY:

select store, date, value
from yourtable
cross apply
(
  values
    ('1/1/2013', [1/1/2013]),
    ('2/1/2013', [2/1/2013])
) c (date, value)

请参阅SQL Fiddle with Demo.所有版本都会给出以下结果:

See SQL Fiddle with Demo. All versions will give a result of:

|   STORE |     DATE | VALUE |
|---------|----------|-------|
| XYZ INC | 1/1/2013 |  1000 |
| XYZ INC | 2/1/2013 |  2000 |

这篇关于如何将列转置为 sql server 中的行的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！