SQL按星期幾分組

Question

使用SQL Server2012。我有一個具有兩個字段的視圖-機器日期和總分鍾數。 該視圖僅顯示前一周的數據-效果很好。 然后，我使用下面的SQL基於此視圖創建另一個視圖：

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，結果不是一行，它們隨着一周的每一天而增加，即

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

但是我需要結果顯示如下：

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

我要去哪里錯了？

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

關於使用pviot的評論是另一種實現方式：

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

我發現語法有點混亂，但這應該可以工作：

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])部分返回表示星期幾的1到7之間的數字，該數字用於指定[1],[2],[3]...部分中的列。查詢。

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

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