SQL按星期几分组

Question

使用SQL Server2012。我有一个具有两个字段的视图-机器日期和总分钟数。 该视图仅显示前一周的数据-效果很好。 然后，我使用下面的SQL基于此视图创建另一个视图：

SELECT        
    CASE WHEN DATENAME(dw, MachineMidLineDate) = 'Sunday' THEN [Total Mins] END AS Sun, 
    CASE WHEN DATENAME(dw, MachineMidLineDate) = 'Monday' THEN [Total Mins] END AS Mon, 
    CASE WHEN DATENAME(dw, MachineMidLineDate) = 'Tuesday' THEN [Total Mins] END AS Tues, 
    CASE WHEN DATENAME(dw, MachineMidLineDate) = 'Wednesday' THEN [Total Mins] END AS Wed, 
    CASE WHEN DATENAME(dw, MachineMidLineDate) = 'Thursday' THEN [Total Mins] END AS Thursday, 
    CASE WHEN DATENAME(dw, MachineMidLineDate) = 'Friday' THEN [Total Mins] END AS Friday, 
    CASE WHEN DATENAME(dw, MachineMidLineDate) = 'Saturday' THEN [Total Mins] END AS Sat
FROM
    dbo.vw_Machine_Minutes_Overview

但是，结果不是一行，它们随着一周的每一天而增加，即

Sun    Mon   Tues   Wed  Thurs  Fri   Sat
10     
       15
              25
                    20
                          21
                                12

但是我需要结果显示如下：

Sun    Mon   Tues   Wed  Thurs  Fri   Sat
10     15     25    20    21    12

我要去哪里错了？

Answer 1

您要聚合：

SELECT SUM(CASE WHEN DATENAME(dw, MachineMidLineDate) = 'Sunday' THEN [Total Mins] END) AS Sun,
       SUM(CASE WHEN DATENAME(dw, MachineMidLineDate) = 'Monday' THEN [Total Mins] END) AS Mon,
      . . . 
FROM dbo.vw_Machine_Minutes_Overview;

Answer 2

关于使用pviot的评论是另一种实现方式：

工作示例： http : //data.stackexchange.com/stackoverflow/revision/749735/930554/pivot-data-by-day-of-week-from-date-field

我发现语法有点混乱，但这应该可以工作：

select *
from 
(
select  datepart(dw,[MachineMidLineDate]) as DayOfWeek, [Total Mins]
  from #vw_Machine_Minutes_Overview 
) src
pivot
(
  sum([Total Mins])
  for DayOfWeek in ([1], [2], [3],[4],[5],[6],[7])
) piv;

datepart(dw,[MachineMidLineDate])部分返回表示星期几的1到7之间的数字，该数字用于指定[1],[2],[3]...部分中的列。查询。

datepart： https : //docs.microsoft.com/zh-cn/sql/t-sql/functions/datepart-transact-sql

枢纽： https ： //technet.microsoft.com/zh-CN/library/ms177410（v = sql.105）.aspx