所以，我想，对于开始日期和结束日期，确定在这两个日期之间一周中有多少特殊的日子。
那么有多少个星期一、星期二等等
我知道我可以在开始日期和结束日期之间循环并每天检查，但差异可能会有很多天。我更喜欢不需要循环的东西。什么好主意吗?(SQL Server 2005+必须支持)
考虑到我认为你想要得到的东西，这个应该可以做到:
SET DATEFIRST 1
DECLARE
@start_date DATETIME,
@end_date DATETIME
SET @start_date = '2011-07-11'
SET @end_date = '2011-07-22'
;WITH Days_Of_The_Week AS (
SELECT 1 AS day_number, 'Monday' AS day_name UNION ALL
SELECT 2 AS day_number, 'Tuesday' AS day_name UNION ALL
SELECT 3 AS day_number, 'Wednesday' AS day_name UNION ALL
SELECT 4 AS day_number, 'Thursday' AS day_name UNION ALL
SELECT 5 AS day_number, 'Friday' AS day_name UNION ALL
SELECT 6 AS day_number, 'Saturday' AS day_name UNION ALL
SELECT 7 AS day_number, 'Sunday' AS day_name
)
SELECT
day_name,
1 + DATEDIFF(wk, @start_date, @end_date) -
CASE WHEN DATEPART(weekday, @start_date) > day_number THEN 1 ELSE 0 END -
CASE WHEN DATEPART(weekday, @end_date) < day_number THEN 1 ELSE 0 END
FROM
Days_Of_The_Week
我不确定OP之后是什么，这将给出每周每天的计数:
SET DATEFIRST 1
DECLARE @StartDate datetime
,@EndDate datetime
SELECT @StartDate='7/13/2011'
,@EndDate='7/28/2011'
;with AllDates AS
(
SELECT @StartDate AS DateOf, datename(weekday,@StartDate) AS WeekDayName, datepart(weekday,@StartDate) AS WeekDayNumber
UNION ALL
SELECT DateOf+1, datename(weekday,DateOf+1), datepart(weekday,DateOf+1)
FROM AllDates
WHERE DateOf<@EndDate
)
SELECT COUNT(*) CountOf,WeekDayName FROM AllDates GROUP BY WeekDayName,WeekDayNumber ORDER BY WeekDayNumber
CountOf WeekDayName
----------- ------------------------------
2 Monday
2 Tuesday
3 Wednesday
3 Thursday
2 Friday
2 Saturday
2 Sunday
(7 row(s) affected)
这将给出星期一到星期五的天数:
SET DATEFIRST 1
DECLARE @StartDate datetime
,@EndDate datetime
SELECT @StartDate='7/13/2011'
,@EndDate='7/28/2011'
;with AllDates AS
(
SELECT @StartDate AS DateOf, datepart(weekday,@StartDate) AS WeekDayNumber
UNION ALL
SELECT DateOf+1, datepart(weekday,DateOf+1)
FROM AllDates
WHERE DateOf<@EndDate
)
SELECT COUNT(*) AS WeekDayCount FROM AllDates WHERE WeekDayNumber<=5
WeekDayCount
------------
12
(1 row(s) affected)
如果您有一个假日表，您可以加入它并删除那些。这里有一个稍微不同的版本，可能执行得更好:
SET DATEFIRST 1
DECLARE @StartDate datetime
,@EndDate datetime
SELECT @StartDate='7/13/2011'
,@EndDate='7/28/2011'
;with AllDates AS
(
SELECT @StartDate AS DateOf, datepart(weekday,getdate()) AS WeekDayNumber
UNION ALL
SELECT DateOf+1, (WeekDayNumber+1) % 7
FROM AllDates
WHERE DateOf<@EndDate
)
SELECT COUNT(*) AS WeekDayCount FROM AllDates WHERE WeekDayNumber>0 AND WeekDayNumber<6
--I don't like using "BETWEEN", ">", ">=", "<", and "<=" are more explicit in defining end points
产生与原始查询相同的输出。
这是标准设置，但可以调整
DECLARE @StartDate datetime, @EndDate datetime
SELECT @StartDate='20110601', @EndDate='20110630'
;WITH AllDates AS
(
SELECT @StartDate AS DateOf, datepart(weekday, @StartDate) AS WeekDayNumber
UNION ALL
SELECT DateOf+1, datepart(weekday, DateOf+1)
FROM AllDates
WHERE DateOf < @EndDate
)
SELECT SUM(CASE WHEN WeekDayNumber BETWEEN 2 AND 6 THEN 1 ELSE 0 END) AS WeekDayCount
FROM AllDates
OPTION (MAXRECURSION 0)
这应该对SQL Server有效，并且应该是国际化安全的(注意:我做没有有一个服务器来测试它)。
SELECT datediff(day, @start, @end) - datediff(week, @start, @end) * 2
- CASE WHEN datepart(weekday, @start)
IN (datepart(weekday, '1970-01-03'),
datepart(weekday, '1970-01-04'))
THEN 1
ELSE 0 END,
- CASE WHEN datepart(weekday, @end)
IN (datepart(weekday, '1970-01-03'),
datepart(weekday, '1970-01-04'))
THEN 1
ELSE 0 END
试一试。
给出了说明，这应该得到每天的天数。
不使用递归，并且应该是完全国际安全的。您必须根据需要调整包含/排除的开始/结束日期参数(例如，我用来检查这一点的DB2版本排除了开始日期，但包括了结束日期)。
WITH dayOfWeek (name, dayNumber) as (VALUES(dayname(weekday, '1970-01-01'), daypart(weekday, '1970-01-01')),
(dayname(weekday, '1970-01-02'), daypart(weekday, '1970-01-02')),
(dayname(weekday, '1970-01-03'), daypart(weekday, '1970-01-03')),
(dayname(weekday, '1970-01-04'), daypart(weekday, '1970-01-04')),
(dayname(weekday, '1970-01-05'), daypart(weekday, '1970-01-05')),
(dayname(weekday, '1970-01-06'), daypart(weekday, '1970-01-06')),
(dayname(weekday, '1970-01-07'), daypart(weekday, '1970-01-07')))
SELECT name, dayNumber, datediff(weeks, @start, @end)
+ CASE WHEN datepart(weekday, @end) >= dayNumber THEN 1 ELSE 0 END
- CASE WHEN datepart(weekday, @start) >= dayNumber THEN 1 ELSE 0 END
FROM dayOfWeek
有帮助吗?
您可以使用DATEDIFF和DATEPART函数+一些基本的数学来获得所需的结果，而无需循环。
我本可以简单地将其作为评论添加到标记的答案中，但没有足够的"声誉"。不需要硬编码依赖于datefirst的day_number，您可以通过发现一周中的每一天的工作日来设置它:
;WITH Days_Of_The_Week AS (
SELECT DATEPART(dw, '2007-01-01') AS day_number, 'Monday' AS day_name UNION ALL -- 2007-01-01 is a known Monday
SELECT DATEPART(dw, '2007-01-02') AS day_number, 'Tuesday' AS day_name UNION ALL
SELECT DATEPART(dw, '2007-01-03') AS day_number, 'Wednesday' AS day_name UNION ALL
SELECT DATEPART(dw, '2007-01-04') AS day_number, 'Thursday' AS day_name UNION ALL
SELECT DATEPART(dw, '2007-01-05') AS day_number, 'Friday' AS day_name UNION ALL
SELECT DATEPART(dw, '2007-01-06') AS day_number, 'Saturday' AS day_name UNION ALL
SELECT DATEPART(dw, '2007-01-07') AS day_number, 'Sunday' AS day_name
)
@start_date date = '2017-08-11',
@end_date date = '2017-08-27',
@weekday int = 7,
@count int output
As
Begin
Declare @i int = 0
set @count = 0
while(@i <= (select Datediff(Day, @start_date, @end_date)))
begin
if(Dateadd(Day, @i, @start_date) > @end_date)
break
if(Datepart(weekday, Dateadd(Day, @i, @start_date)) = @weekday)
set @count += 1
set @i += 1
end
select @count