SQL результаты, сгруппированные по неделям, день

Question

У меня есть таблица, как показано ниже:

Date  ServingTime MeasuredValue 
---------------------------------------- 
01-11-16 Breakfast  45 
01-11-16 Lunch   50 
01-11-16 Dinner  60 
02-11-16 Breakfast  23 
02-11-16 Lunch   45 
02-11-16 Dinner  83 
…  
…  
30-11-16 Breakfast  56 
30-11-16 Lunch   16 
30-11-16 Dinner  34 

И запрос, как показано ниже:

DECLARE @DatePeriod date = '2016-11-30' 

SELECT 
    ServingTime, 
    ISNULL([1], 0) AS 'Week 1', 
    ISNULL([2], 0) AS 'Week 2', 
    ISNULL([3], 0) AS 'Week 3', 
    ISNULL([4], 0) AS 'Week 4', 
    ISNULL([5], 0) AS 'Week 5' 

FROM (SELECT 
    ServingTime, 
    DATEDIFF(WEEK, DATEADD(MONTH, DATEDIFF(MONTH, 0, Date), 0), Date) + 1 AS [Weeks], 
    MeasuredValue AS 'Sale' 

FROM HR.FoodCollectionEntry 
-- Only get rows where the date is the same as the DatePeriod 
-- i.e DatePeriod is 30th May 2011 then only the weeks of May will be calculated 
WHERE DATEPART(MONTH, Date) = DATEPART(MONTH, @DatePeriod)) p 
PIVOT (SUM(Sale) FOR Weeks IN ([1], [2], [3], [4], [5], [6])) AS pv 

Этот вывод запроса, как показано ниже:

ServingTime Week 1 Week 2 Week 3 Week 4  Week 5 
--------------------------------------------------------------- 
Breakfast  412  590  510  456  200 
Dinner   329  525  529  529  321 
Lunch   371  529  542  480  233 

Но Я бы привел День также на выходе, чтобы сравнить продажи воскресенья в каждую неделю. Как запросить вывод, как показано ниже?

Day  ServingTime Week 1 Week 2 Week 3 Week 4 Week 5 
--------------------------------------------------------------- 
Sunday  Breakfast 412  590  510  456  200 
Sunday  Dinner  329  525  529  529  321 
Sunday  Lunch  371  529  542  480  233 
Monday  Breakfast 412  590  510  456  200 
Monday  Dinner  329  525  529  529  321 
Monday  Lunch  371  529  542  480  233 
…      
…      
Saturday Breakfast 412  590  510  456  200 
Saturday Dinner  329  525  529  529  321 
Saturday Lunch  371  529  542  480  233 

Answer 1

Просто используйте условную агрегацию:

select datename(weekday, fce.date) as dayOfWeek, servingTime, 
     sum(case when weeks = 1 then sale else 0 end) as Week_1, 
     sum(case when weeks = 2 then sale else 0 end) as Week_2, 
     sum(case when weeks = 3 then sale else 0 end) as Week_3, 
     sum(case when weeks = 4 then sale else 0 end) as Week_4, 
     sum(case when weeks = 5 then sale else 0 end) as Week_5 
from (select fce.*, 
      datediff(week, dateadd(month, datediff(month, 0, Date), 0), Date) + 1 AS Weeks 
     from HR.FoodCollectionEntry fce 
    ) fce 
where month(Date) = month(@DatePeriod) and 
     year(Date) = year(@DatePeriod) -- I assume you want this too 
     group by datename(weekday, fce.date) as dayOfWeek, servingTime 
order by max(datepart(weekday, fce.date)