MSSQL Totals每天一个月

Question

我正在尝试做一些与我之前提出的问题非常类似的事情,但我似乎无法让它正常工作.这是我之前的问题:如何获得每天的总数

该表如下所示:

              Table Name: Totals
Date       |Program label   |count
           |                |    
2013-04-09 |Salary Day      |4364
2013-04-09 |Monthly         |6231
2013-04-09 |Policy          |3523
2013-04-09 |Worst Record    |1423
2013-04-10 |Salary Day      |9872
2013-04-10 |Monthly         |6543
2013-04-10 |Policy          |5324
2013-04-10 |Worst Record    |5432
2013-04-10 |Salary Day      |1245
2013-04-10 |Monthly         |6345
2013-04-10 |Policy          |5431
2013-04-10 |Worst Record    |5232

我的问题是:使用MSSQL 2008 - 有没有办法让我获得当前每月每个程序标签的总计数.正如您所看到的,它有时会每天运行两次.我需要能够解释这一点.

输出应如下所示:

Date      |Salary Day |Monthly |Policy |Worst Record
2013-04-9 |23456      |63241   |23521  |23524
2013-04-10|45321      |72535   |12435  |83612

Answer 1

PIVOT像这样使用表运算符:

SELECT  *
FROM Totals AS t
PIVOT
(
   SUM(count)
   FOR [Program label] IN ([Salary Day], 
                           [Monthly], 
                           [Policy], 
                           [Worst Record])
) AS p;

看到它的实际效果:

• SQL小提琴演示

这会给你:

|       DATE | SALARY DAY | MONTHLY | POLICY | WORST RECORD |
-------------------------------------------------------------
| 2013-04-09 |       4364 |    6231 |   3523 |         1423 |
| 2013-04-10 |      11117 |   12888 |  10755 |        10664 |

Answer 2

尝试这个

select Date,
  sum(case when [Program label] = 'Salary Day' then count else 0 end) [Salary Day],
  sum(case when [Program label] = 'Monthly' then count else 0 end) [Monthly],
  sum(case when [Program label] = 'Policy' then  count else 0 end) [Policy],
  sum(case when [Program label] = 'Worst Record' then count else 0 end) [Worst Record]
from Totals Group by [Date];