计算某个时间在某个日期范围内出现的次数答案

【问题标题】：Count how many times a time occurs within a date range计算某个时间在某个日期范围内出现的次数
【发布时间】：2017-09-15 09:17:09
【问题描述】：

如果已经有人问过这个问题，我想不出一个好的方法来正确搜索这个问题。

我正在 SQL 2008 R2 中寻找一种方法来计算两个日期时间值之间出现了多少次下午 6 点。

例如在 '2017-04-17 19:00:00' 和 '2017-04-19 17:00:00' 6pm 之间仅发生一次虽然时间跨度不同的 3 天。

在 '2017-04-17 18:00:00' 和 '2017-04-19 18:00:00' 之间发生 3 次，同时也跨越3 天。

这是一个非常愚蠢的表达我想要的插图。

timecount(hh, 6, min(datefield), max(datefield))

谢谢

【问题讨论】：

如果你的日期范围很长，比如月份或年份呢？

标签： sql sql-server sql-server-2008 datetime

【解决方案1】：

这里给出任何日期范围之间的计数

declare @time datetime='06:00:00'

declare @startDate datetime='04/20/2017 05:00:00'

declare @enddate datetime='04/21/2017 05:00:00'

SELECT 
  case 
    WHEN datediff(ss,@time, convert(time(0),@startDate)) <=0  and   datediff(ss,@time, convert(time(0),@enddate)) >=0

     THEN datediff(dd,@startDate,@enddate) +1 
WHEN   (datediff(ss,@time, convert(time(0),@startDate)) <=0  and 
datediff(ss,@time, convert(time(0),@enddate)) <=0

        OR
        datediff(ss,@time, convert(time(0),@startDate))> 0  and 

datediff(ss,@time, convert(time(0),@enddate)) >0

        OR

        datediff(ss,@time, convert(time(0),@startDate))> 0  and datediff(ss,@time, convert(time(0),@enddate)) <=0
        )
then  datediff(dd,@startDate,@enddate) 
ELSE  
datediff(dd,@startDate,@enddate)-1
END

【讨论】：

【解决方案2】：

这是两个日期时间之间每下午 6 点的计数：

DECLARE @StartDate datetime
DECLARE @EndDate datetime
set @StartDate = '2017-04-17 19:00:00'
set @EndDate = '2017-04-19 17:00:00'

;WITH cte1 (S) AS (
SELECT 1 FROM (VALUES (1), (1), (1), (1), (1), (1), (1), (1), (1), (1)) n (S)
),
cte2 (S) AS (SELECT 1 FROM cte1 AS cte1 CROSS JOIN cte1 AS cte2),
cte3 (S) AS (SELECT 1 FROM cte1 AS cte1 CROSS JOIN cte2 AS cte2)

select count(datepart(hour,result)) as count from
(SELECT TOP (DATEDIFF(hour, @StartDate, @EndDate) + 1)
        result = DATEADD(hour, ROW_NUMBER() OVER(ORDER BY S) - 1, @StartDate)
FROM cte3) as res where datepart(hour,result) = 18

这是两个日期时间之间下午 6 点的详细视图：

DECLARE @StartDate datetime
DECLARE @EndDate datetime
set @StartDate = '2017-04-17 19:00:00'
set @EndDate = '2017-04-19 17:00:00'

;WITH cte1 (S) AS (
SELECT 1 FROM (VALUES (1), (1), (1), (1), (1), (1), (1), (1), (1), (1)) n (S)
),
cte2 (S) AS (SELECT 1 FROM cte1 AS cte1 CROSS JOIN cte1 AS cte2),
cte3 (S) AS (SELECT 1 FROM cte1 AS cte1 CROSS JOIN cte2 AS cte2)

select result,datepart(hour,result)  from
(SELECT TOP (DATEDIFF(hour, @StartDate, @EndDate) + 1)
        result = DATEADD(hour, ROW_NUMBER() OVER(ORDER BY S) - 1, @StartDate)
FROM cte3) as res where datepart(hour,result) = 18

【讨论】：

【解决方案3】：

试试下面的脚本，使用CTE

DECLARE @F_DATE     AS DATETIME =   '2017-04-17 19:00:00'
        ,@T_DATE    AS DATETIME =   '2017-04-19 17:00:00'


;WITH CTE
AS  (
    SELECT  (CASE WHEN DATEPART(HH,@F_DATE) <= 18
                THEN @F_DATE 
                ELSE (CASE WHEN  DATEDIFF(DAY,@F_DATE,@T_DATE) > 0 
                        THEN DATEADD(DAY,1,@F_DATE) END)
             END)   AS  CDATE

    UNION ALL

    SELECT  DATEADD(DAY,1,CDATE)
    FROM    CTE
    WHERE   DATEADD(DAY,1,CDATE)    <=  @T_DATE
)

SELECT COUNT(CDATE) DATE_COUNT 
FROM CTE
OPTION ( MAXRECURSION 0 )

【讨论】：

CTE 有一些流程：最大递归深度和性能。除非没有其他办法，否则最好避免使用 CTE。

【解决方案4】：

试试下面的公式，我尝试过不同的场景，它可以工作，如果我错过任何场景并且不能按照您的要求工作，请告诉我。

DECLARE @firstDate Datetime='17-Apr-2017 17:00:00'
DECLARE @secondDate Datetime='17-Apr-2017 18:59:00'


SELECT
    CASE WHEN DATEDIFF(day,@firstDate,@secondDate)=0
    THEN IIF(CAST(@firstDate AS TIME) <='18:00' AND DATEPART(hh,@secondDate) >=18,1,0)
    ELSE 
        CASE WHEN 
            (
                CAST(@firstDate AS TIME) <='18:00' AND CAST(@secondDate AS TIME) <'18:00' 
                OR
                CAST(@firstDate AS TIME) >'18:00' AND CAST(@secondDate AS TIME) >='18:00' 
            )
            THEN DATEDIFF(day,@firstDate,@secondDate)
            WHEN CAST(@firstDate AS TIME) <='18:00' AND CAST(@secondDate AS TIME) >='18:00' THEN DATEDIFF(day,@firstDate,@secondDate)+1 
            ELSE DATEDIFF(day,@firstDate,@secondDate)-1
            END
    END AS TotalCount

【讨论】：

嗨，我遇到了语法错误，但无法弄清楚它是什么 Msg 102，Level 15，State 1，Line 7 '
不可能，使用我的 SQL，我只是再次测试
一切皆有可能 :)，也许我不知道不同的 SQL 版本。
:-) 这可能是个问题，无论您以何种方式得到答案
我认为不是服务器问题，因为您显示的错误与版本无关

【解决方案5】：

一个简单的计数查询：

DECLARE @StartDate datetime = '2017-04-17 18:00:00'

DECLARE @EndDate datetime = '2017-04-19 18:00:00'

SELECT 
   CASE 
      WHEN CAST(@StartDate AS time) <= '18:00' AND CAST(@EndDate AS time) >= '18:00' 
             THEN datediff(day, @StartDate, @EndDate) + 1
      WHEN CAST(@StartDate AS time) <= '18:00' AND CAST(@EndDate AS time) < '18:00' 
             THEN datediff(day, @StartDate, @EndDate)      
      WHEN CAST(@StartDate AS time) > '18:00' AND CAST(@EndDate AS time) >= '18:00' 
             THEN datediff(day, @StartDate, @EndDate)
      ELSE datediff(day, @StartDate, @EndDate) - 1
   END AS TotalCount

【讨论】：

谢谢，这工作做得很好。

【解决方案6】：

DECLARE 
    @Time time = '18:00',
    @From datetime = '2017-04-17 18:00:00',
    @To datetime = '2017-04-19 18:00:00'


SELECT 
    CASE 
        -- Same date
        WHEN DATEDIFF(DAY, @From, @To) = 0 THEN 
            CASE WHEN CAST(CAST(@From AS date) AS datetime) + @Time BETWEEN @From AND @To THEN 1 ELSE 0 END
        -- Not same date
        WHEN @From <= @To THEN
            CASE WHEN @Time >= CAST(@From AS time) THEN 1 ELSE 0 END 
            + DATEDIFF(DAY, @From, @To) - 1
            + CASE WHEN @Time <= CAST(@To AS time) THEN 1 ELSE 0 END 
        -- Invalid range
        ELSE 0
    END AS CountOfTime

【讨论】：

【解决方案7】：

这将为您提供每小时和发生次数：

select datepart(hh, DateColumn) as TheHours, count(*) as occurs
from MyTable
where DateColumn between @SomeDate and @SomeOtherDate
group by datepart(hh, DateColumn)

或者只是下午 6 点：

select count(*)
from MyTable
where datepart(hh, DateColumn) = 18
and DateColumn between @SomeDate and @SomeOtherDate

【讨论】：