問題描述
我正在嘗試獲取一個原始數據集，為新數據添加列并將其轉換為更傳統的表結構.這個想法是讓腳本提取列名稱(日期)并將其放入一個新列中，然后將每個日期數據值堆疊在一起.
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
謝謝
推薦答案
有幾種不同的方法可以獲得您想要的結果.
There are a few different ways that you can get the result that you want.
您可以將 SELECT 與 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;
參見SQL Fiddle with Demo.
您可以使用 UNPIVOT 函數:
You can use the UNPIVOT function:
select store, date, value
from yourtable
unpivot
(
value
for date in ([1/1/2013], [2/1/2013])
) un;
參見 SQL Fiddle with Demo.
最后，根據您的 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)
請參閱SQL Fiddle with Demo.所有版本都會給出以下結果:
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 |
這篇關于如何將列轉置為 sql server 中的行的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！