Question

我需要比较2个日期并返回两者之间的天数。这是一个表格示例：

+----+--------+-------------------------+-------------------------+
| id | userid |       datestarted       |     datefinished        |
+----+--------+-------------------------+-------------------------+
|    |        |                         |                         |
|  1 |     23 | 2014-03-25 09:05:00.000 | 2014-03-25 12:15:00.000 |
|  2 |     43 | 2014-03-25 09:05:00.000 | 2014-03-25 12:15:00.000 |
|  3 |     23 | 2014-03-31 09:05:00.000 | 2014-03-31 12:15:00.000 |
|  4 |     12 | 2014-03-25 09:05:00.000 | 2014-03-26 12:15:00.000 |
+----+--------+-------------------------+-------------------------+

在前3个案例中，我们在同一天，只有几个小时不匹配。

Datestarted = 2014-03-25 09:05:00.000

Datefinished = 2014-03-25 12:15:00.000

我们只输入hours和minutes。

到目前为止，我们只需将difference显示为whole number，而不是decimal points，并且这样做：

DATEDIFF(carsharing.datestarted, carsharing.datefinished)

但是现在，我们必须在difference之间显示0,5 day，如果它是less而不是4,5 hours。如果差异为greater，则应保持为1 day。

在表中更复杂的最后一个案例中，我们还应该比较并显示两个不同日期之间的差异

Datestarted = 2014-03-25 09:05:00.000

Datefinished = 2014-03-26 12:15:00.000

此处的结果应为1,5 days

Answer 1

我相信这是你正在寻找的东西 - 这将把当天4.5小时以内的任何东西的差异缩小到0.5，而其他所有东西都会花一整天：

Declare @StartDate  DateTime = '2014-03-25 09:05:00.000',
        @EndDate    DateTime = '2014-03-26 12:15:00.000'

;With TotalHours As
(
    Select  DateDiff(Minute, @StartDate, @EndDate) / 60.0 As TotalHours
)
Select  Case 
            When TotalHours % 24 = 0
                Then Floor(TotalHours / 24)
            When TotalHours % 24 < 4.5 
                Then Floor(TotalHours / 24) + 0.5 
            Else Floor(TotalHours / 24) + 1.0
        End As Days
From    TotalHours

Answer 2

DECLARE 
    @StartDate datetime = '2014-03-25 09:05:00.000'
    ,@EndDate datetime = '2014-03-26 09:05:00.000'
;WITH d AS (SELECT DATEDIFF(d,@StartDate,@EndDate) Dys) 
,h AS (SELECT DATEDIFF(hh,@StartDate,@EndDate) Hrs)
SELECT d.Dys + CASE WHEN (h.Hrs - d.Dys*24) = 0 THEN 0 ELSE CASE WHEN (h.Hrs - d.Dys*24) < 4.5 THEN 0.5 ELSE 1 END END
FROM d,h

Answer 3

您可以尝试此查询。它以分钟为单位得到差异并乘以2以获得0.5天的范围。然后在计算天花板值之前将其分为24小时60分钟。一旦你拥有它，它可以再次偏离2。

当价值超过4.5 * 24 * 60（4.5天（分钟））时，只需要分成24和60.

<强>查询：

Select id, userid, datestarted, datefinished
    , Days = Case When DATEDIFF(minute, datestarted, datefinished) > 4.5*60*24      
        then DATEDIFF(minute, datestarted, datefinished) / 24 / 60
        else CEILING(((2.0*DATEDIFF(minute, datestarted, datefinished)) / 24 / 60)) / 2
    end
From @dates

<强>输出：

id  userid  datestarted                 datefinished                Days
1   23      2014-03-25 09:05:00.000     2014-03-25 12:15:00.000     0.500000
2   43      2014-03-25 09:05:00.000     2014-03-25 12:15:00.000     0.500000
3   23      2014-03-31 09:05:00.000     2014-03-31 12:15:00.000     0.500000
4   12      2014-03-25 09:05:00.000     2014-03-26 12:15:00.000     1.500000
5   12      2014-03-25 09:05:00.000     2014-03-29 12:15:00.000     4.500000
6   12      2014-03-25 09:05:00.000     2014-03-29 22:15:00.000     4.000000

示例数据

declare @dates table(id int, userid int, datestarted datetime, datefinished datetime);
insert into @dates(id, userid, datestarted, datefinished) values
(1, 23, '2014-03-25 09:05:00.000', '2014-03-25 12:15:00.000')
, (2, 43, '2014-03-25 09:05:00.000', '2014-03-25 12:15:00.000')
, (3, 23, '2014-03-31 09:05:00.000', '2014-03-31 12:15:00.000')
, (4, 12, '2014-03-25 09:05:00.000', '2014-03-26 12:15:00.000')
, (5, 12, '2014-03-25 09:05:00.000', '2014-03-29 12:15:00.000')
, (6, 12, '2014-03-25 09:05:00.000', '2014-03-29 22:15:00.000')

SQL计算Datediff为8个工作日的一半

3 个答案: