Datediff和枢轴

Question

我有这张桌子（实际桌子要大得多）：

id     class     startdate      enddate    
----------------------------------------
1       High       1/1/15       2/1/15      
1       Low        5/1/15       6/1/15      
1       Mid        6/1/15       6/10/15   

有没有办法产生如下结果...

 id  High    Mid    low
 -----------------------
 1    30      9      30    

我的工作如下......（只是表明我正在尝试解决这个问题）

SELECT DISTINCT   --DATEDIFF(DAY, tt.START_DATE,tt.END_DATE)AS Stay_days, 
    tt.[high], tt.[mid], tt.[low], tt.id,
FROM            
  (SELECT 
       [pvt_High], [pvt_mid], [pvt_low], pvt.STARTDATE, pvt.ENDDATE, pvt.id
   FROM 
       (SELECT DISTINCT   
            l.id, L.class, l.STARTDATE, l.ENDDATE,
            DATEDIFF(DAY, l.STARTDATE, l.ENDDATE) AS STAY_DAYS,
            'Y' AS Flag
        FROM TEST l
        WHERE 1=1

        PIVOT (MAX(Flag) FOR CLASS IN ([pvt_HIGH], [pvt_MID],[pvt_LOW]))
        AS pvt
    ) tt

Answer 1

您可以使用内部选择将数据转换为您拥有ID，类和天数的表单。然后，您可以将类值上的数据转换为列。

select *
from
(
    select id, class, datediff(day, startDate, endDate) as days
    from @test
) x
pivot
(
    max(days) for class in (high, low, med)
) y

SQL Fiddle

显然，将名称等更改为适合您数据的内容，这只是一个示例。