我有这张桌子：
Vacationtbl ：
ID Start End
-------------------------
01 04/10/17 04/12/17
01 04/27/17 05/02/17
02 04/13/17 04/15/17
02 04/17/17 04/20/17
03 06/14/17 06/22/17
Employeetbl:
ID Fname Lname
------------------
01 John AAA
02 Jeny BBB
03 Jeby CCC
我喜欢计算每位员工4月份休假的天数。
我的查询：
SELECT
SUM(DATEDIFF(DAY, Start, End) + 1) AS Days
FROM
Vacationtbl
GROUP BY
ID
01返回9（不正确）02返回7（正确）如何修复查询，使其计算到月末，并在月末停止。例如，4月有30天。在第二行，员工01应在4/27/17之前计算4/30/17。 05/02/17适用于五月。
由于
答案 0 :(得分：3)
Tally / Calendar表是最佳选择。但是，您可以使用 ad-hoc 计数表。
示例
Select Year = Year(D)
,Month = Month(D)
,ID
,Days = count(*)
From Vacationtbl A
Cross Apply (
Select Top (DateDiff(DAY,[Start],[End])+1) D=DateAdd(DAY,-1+Row_Number() Over (Order By (Select Null)),[Start])
From master..spt_values
) B
-- YOUR OPTIONAL WHERE STATEMENT HERE --
Group By ID,Year(D),Month(D)
Order By 1,2,3
<强>返回
Year Month ID Days
2017 4 01 7
2017 4 02 7
2017 5 01 2
编辑 - 即使零日显示所有ID
Select ID
,Year = Year(D)
,Month = Month(D)
,Days = sum(case when D between [Start] and [End] then 1 else 0 end)
From (
Select Top (DateDiff(DAY,'05/01/2017','05/31/2017')+1) D=DateAdd(DAY,-1+Row_Number() Over (Order By (Select Null)),'05/01/2017')
From master..spt_values
) D
Cross Join Vacationtbl B
Group By ID,Year(D),Month(D)
Order By 1,2,3
<强>返回
ID Year Month Days
1 2017 5 2
2 2017 5 0
编辑 - 2纠正重叠（差距和群岛）
--Create Some Sample Data
----------------------------------------------------------------------
Declare @Vacationtbl Table ([ID] varchar(50),[Start] date,[End] date)
Insert Into @Vacationtbl Values
(01,'04/10/17','04/12/17')
,(01,'04/27/17','05/02/17')
,(02,'04/13/17','04/15/17')
,(02,'04/17/17','04/20/17')
,(02,'04/16/17','04/17/17') -- << Overlap
,(03,'05/16/17','05/17/17')
-- The Actual Query
----------------------------------------------------------------------
Select ID
,Year = Year(D)
,Month = Month(D)
,Days = sum(case when D between [Start] and [End] then 1 else 0 end)
From (Select Top (DateDiff(DAY,'04/01/2017','04/30/2017')+1) D=DateAdd(DAY,-1+Row_Number() Over (Order By (Select Null)),'04/01/2017') From master..spt_values ) D
Cross Join (
Select ID,[Start] = min(D),[End] = max(D)
From (
Select E.*,Grp = Dense_Rank() over (Order By D) - Row_Number() over (Partition By ID Order By D)
From (
Select Distinct A.ID,D
From @Vacationtbl A
Cross Apply (Select Top (DateDiff(DAY,A.[Start],A.[End])+1) D=DateAdd(DAY,-1+Row_Number() Over (Order By (Select Null)),A.[Start]) From master..spt_values ) B
) E
) G
Group By ID,Grp
) B
Group By ID,Year(D),Month(D)
Order By 1,2,3
<强>返回
ID Year Month Days
1 2017 4 7
2 2017 4 8
3 2017 4 0
答案 1 :(得分：1)
没有日期表，您可以使用
select Id
,sum(case when [end]>'20170430' and [start]<'20170401' then datediff(day,'20170401','20170430')+1
when [end]>'20170430' then datediff(day,[start],'20170430')+1
when [start]<'20170401' then datediff(day,'20170401',[end])+1
else datediff(day,[start],[end])+1
end) as VacationDays
from Vacationtbl
where [start] <= '20170430' and [end] >= '20170401'
group by Id
这里有3个条件
编辑：根据OP的评论，未来的日期必须包括在内，
/*This recursive cte generates the month start and end dates with in a given time frame
For Eg: all the month start and end dates for 2017
Change the start and end period as needed*/
with dates (month_start_date,month_end_date) as
(select cast('2017-01-01' as date),cast(eomonth('2017-01-01') as date)
union all
select dateadd(month,1,month_start_date),eomonth(dateadd(month,1,month_start_date)) from dates
where month_start_date < '2017-12-01'
)
--End recursive cte
--Query logic is the same as above
select v.Id
,year(d.month_start_date) as yr,month(d.month_start_date) as mth
,sum(case when v.[end]>d.month_end_date and v.[start]<d.month_start_date then datediff(day,d.month_start_date,d.month_end_date)+1
when v.[end]>d.month_end_date then datediff(day,v.[start],d.month_end_date)+1
when v.[start]<d.month_start_date then datediff(day,d.month_start_date,v.[end])+1
else datediff(day,v.[start],v.[end])+1
end) as VacationDays
from dates d
join Vacationtbl v on v.[start] <= d.month_end_date and v.[end] >= d.month_start_date
group by v.id,year(d.month_start_date),month(d.month_start_date)
答案 2 :(得分：0)
您可以使用日历或日期表进行此类操作。
对于内存中只有152kb，您可以在表格中拥有30年的日期：
/* dates table */
declare @fromdate date = '20000101';
declare @years int = 30;
/* 30 years, 19 used data pages ~152kb in memory, ~264kb on disk */
;with n as (select n from (values(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) t(n))
select top (datediff(day, @fromdate,dateadd(year,@years,@fromdate)))
[Date]=convert(date,dateadd(day,row_number() over(order by (select 1))-1,@fromdate))
into dbo.Dates
from n as deka cross join n as hecto cross join n as kilo
cross join n as tenK cross join n as hundredK
order by [Date];
create unique clustered index ix_dbo_Dates_date
on dbo.Dates([Date]);
如果不采取创建表格的实际步骤，您可以在common table expression内使用它：
declare @fromdate date = '20170401';
declare @thrudate date = '20170430';
;with n as (select n from (values(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) t(n))
, dates as (
select top (datediff(day, @fromdate, @thrudate)+1)
[Date]=convert(date,dateadd(day,row_number() over(order by (select 1))-1,@fromdate))
from n as deka cross join n as hecto cross join n as kilo
cross join n as tenK cross join n as hundredK
order by [Date]
)
select [Date]
from dates;
使用如下：
select
v.Id
, count(*) as VacationDays
from Vacationtbl v
inner join Dates d
on d.Date >= v.[Start]
and d.Date <= v.[End]
where d.Date >= '20170401'
and d.Date <= '20170430'
group by v.Id
rextester demo（表格）：http://rextester.com/PLW73242
rextester demo（cte）：http://rextester.com/BCY62752
返回：
+----+--------------+
| Id | VacationDays |
+----+--------------+
| 01 | 7 |
| 02 | 7 |
+----+--------------+
数字和日历表参考：
答案 3 :(得分：0)
假设您只想要一个月并且想要计算所有日期，您可以使用算术来完成此操作。不需要单独的日历表。优势在于性能。
我认为如果SQL Server支持least()和greatest()会更容易，但case会这样做：
select id,
sum(1 + datediff(day, news, newe)) as vacation_days_april
from vactiontbl v cross apply
(values (case when [start] < '2017-04-01' then cast('2017-04-01' as date) else [start] end),
(case when [end] >= '2017-05-01' then cast('2017-04-30' as date) else [end] end)
) v(news, newe)
where news <= newe
group by id;
您可以随时将其扩展到任何月份：
with m as (
select cast('2017-04-01' as date) as month_start,
cast('2017-04-30' as date) as month_end
)
select id,
sum(1 + datediff(day, news, newe)) as vacation_days_aprile
from m cross join
vactiontbl v cross apply
(values (case when [start] < m.month_start then m.month_start else [start] end),
(case when [end] >= m.month_end then m.month_end else [end] end)
) v(news, newe)
where news <= newe
group by id;
您甚至可以使用类似的想法延长到多个月，每个用户和每个月都有不同的行。
答案 4 :(得分：0)
试试这个，
declare @Vacationtbl table(ID int,Startdate date,Enddate date)
insert into @Vacationtbl VALUES
(1 ,'04/10/17','04/12/17')
,(1 ,'04/27/17','05/02/17')
,(2 ,'04/13/17','04/15/17')
,(2 ,'04/17/17','04/20/17')
-- somehow convert your input into first day of month
Declare @firstDayofGivenMonth date='2017-04-01'
Declare @LasttDayofGivenMonth date=dateadd(day,-1,dateadd(month,datediff(month,0,@firstDayofGivenMonth)+1,0))
;with CTE as
(
select *
,case when Startdate<@firstDayofGivenMonth then @firstDayofGivenMonth else Startdate end NewStDT
,case when Enddate>@LasttDayofGivenMonth then @LasttDayofGivenMonth else Enddate end NewEDT
from @Vacationtbl
)
SELECT
SUM(DATEDIFF(DAY, NewStDT, NewEDT) + 1) AS Days
FROM
CTE
GROUP BY
ID