Question

使用SQL Server 2008我想计算在给定日期范围内16:30到18:00之间的分钟数。

因此输入例如@fromdate = ＆＃39; 2017-04-17 17：00：00＆＃39; 和@todate = ＆＃39; 2017-04- 19 17：00：00＆＃39;

所以对于上面使用拉门数学的例子。

Day 1 (17th) : 60
Day 2 (18th) : 90
Day 3 (19th) : 30

Total 180

第1天是60，因为从开始日期17:00到18:00之间有60分钟。第2天是90，因为中间日期必须是16:30到18:00之间的完整90分钟，第3天是30在16:30到17:00之间有30分钟（结束日期的时间）所以如果有意义的话，我每天都会在16:30到18:00之间搜索分钟数。

我希望总数是输出。

Answer 1

这将在sql server中运行。我们在cte中填充开始和结束日期时间。在那之后，它就像在几分钟内总结日期差异一样简单：

declare @fromdate datetime = convert(datetime,'2017-04-17 17:00:00',20)
declare @todate datetime = convert(datetime,'2017-04-19 17:00:00',20)

declare @startRangeTime time = '16:30';
declare @endRangeTime time = '18:00';

with dates as (
    select top (datediff(day, @fromdate, @toDate) + 1)
    case when dateadd(day,row_number() over  (order by a.object_id) - 1, cast(@fromdate as date)) = cast(@fromdate as date)
        then @fromdate
        else cast(dateadd(day,row_number() over  (order by a.object_id) - 1, cast(@fromdate as date)) as datetime) + cast(@startRangeTime as datetime)
    end as startDt,
    case when dateadd(day,row_number() over  (order by a.object_id) - 1, cast(@fromdate as date)) = cast(@todate as date)
        then @todate
        else cast(dateadd(day,row_number() over  (order by a.object_id) - 1, cast(@fromdate as date)) as datetime) + cast(@endRangeTime as datetime)
    end as endDt
    from sys.all_objects a
    cross join sys.all_objects b
)

select sum(datediff(mi, startDt, endDt))  from dates

返回：

180

在generating a sequence without loops

上生成没有循环的日期

Answer 2

你不能减去2个字符串，所以这样做。用于获取2个日期之间的分钟时间的示例代码。

ALTER SESSION SET NLS_LANGUAGE="American" 
WITH source as (
    SELECT '30-OCT-2016 21:08:34' as A, '30/10/2016 21:06:34' as B
    FROM Dual
)
SELECT TO_DATE(A, 'DD-MON-YYYY HH24:MI:SS') as new_a,
   TO_DATE(B, 'DD/MM/YYYY HH24:MI:SS') as new_b,
   (  TO_DATE(A, 'DD-MON-YYYY HH24:MI:SS') 
    - TO_DATE(B, 'DD/MM/YYYY HH24:MI:SS')
   ) * 24 * 60 as result
FROM source

希望这会有所帮助......

为日期范围计算两次之间的分钟数

2 个答案: