0

我有这种类型的桌子

CREATE TABLE attendance
(
  Date datetime,
  Timein datetime,
  Timeout datetime,
  Spend nvarchar(50),
  excessshort nvarchar(50) 
)

我的数据看起来像这样。

Date                     TimeIn                   Timeout                 Spend     excessshort
2013-01-01 00:00:00.000  2013-01-01 09:14:00.000  2013-01-01 19:06:00.000 09:52:00

我想计算超额

多余的短来自花费的时间意味着员工花费了 9 小时 52 分钟，然后是多余的 00:52 分钟，如果员工花费了 8 小时 44 分钟，那么其短的 00:16 分钟我们有 9 到 6 小时的定时班次。

4

1 回答 1

0

这是一个快速而肮脏的解决方案。简而言之，该支出列没有任何帮助，因为它是一个 varchar，我们需要进行日期数学运算。

出于演示目的，我用我需要的三列创建了一个表变量。使用 datediff 函数，我们将计算每个人打卡的分钟数，然后减去 9 小时（540 分钟）中的分钟数。

如果答案是肯定的，那么这个人的打卡时间超过了 9 个小时。如果是负数，他们就达不到要求了。如果正好是 540 分钟，那么他们就准时了。

就像我说的，它又快又脏。您必须根据您的具体情况调整这些概念。我不知道您是要更新表格还是只是运行报告或其他内容。

declare @attendance TABLE
(
    id int IDENTITY(1,1),
    Date datetime,
    Timein datetime,
    Timeout datetime
)
insert into @attendance (date,timein,timeout) values ('2013-01-01 00:00:00.000','2013-01-01 09:14:00.000','2013-01-01 19:06:00.000')
insert into @attendance (date,timein,timeout) values ('2013-01-01 00:00:00.000','2013-01-01 09:14:00.000','2013-01-01 17:45:00.000')
insert into @attendance (date,timein,timeout) values ('2013-01-01 00:00:00.000','2013-01-01 09:00:00.000','2013-01-01 18:00:00.000')


declare @shiftHours int, @shiftMinutes int
set @shiftHours = 9
set @shiftMinutes = @shiftHours * 60


select 
    datediff(minute,timein,timeout) as minutesClockedIn, 
    datediff(minute,timein,timeout) - @shiftMinutes as offset,
    excessOrShort =
        case 
            when (datediff(minute,timein,timeout) - @shiftMinutes) > 0 then 'excess'
            when (datediff(minute,timein,timeout) - @shiftMinutes) < 0 then 'short'
            else 'ontime'
        end,
    excessOrShortDescription =
        case 
            when (datediff(minute,timein,timeout) - @shiftMinutes) > 0
            then right('0' + cast(abs((datediff(minute,timein,timeout) - @shiftMinutes) / 60) as varchar(2)),2) + ':' + right('0' + cast(abs((datediff(minute,timein,timeout) - @shiftMinutes) % 60) as varchar(2)),2) + ' in excess'
            when (datediff(minute,timein,timeout) - @shiftMinutes) < 0
            then right('0' + cast(abs((datediff(minute,timein,timeout) - @shiftMinutes) / 60) as varchar(2)),2) + ':' + right('0' + cast(abs((datediff(minute,timein,timeout) - @shiftMinutes) % 60) as varchar(2)),2) + ' short'
            else 'ontime'
        end

from @attendance

下面的这个 CTE 版本更具可读性，但做同样的事情。我只是使用 CTE 来消除一些重复的日期差异

declare @attendance TABLE
(
    id int IDENTITY(1,1),
    Date datetime,
    Timein datetime,
    Timeout datetime
)
insert into @attendance (date,timein,timeout) values ('2013-01-01 00:00:00.000','2013-01-01 09:14:00.000','2013-01-01 19:06:00.000')
insert into @attendance (date,timein,timeout) values ('2013-01-01 00:00:00.000','2013-01-01 09:14:00.000','2013-01-01 17:45:00.000')
insert into @attendance (date,timein,timeout) values ('2013-01-01 00:00:00.000','2013-01-01 09:00:00.000','2013-01-01 18:00:00.000')


declare @shiftHours int, @shiftMinutes int
set @shiftHours = 9
set @shiftMinutes = @shiftHours * 60
-- populate CTE for use by select statement below
;with shiftData as (
    select
        date,
        datediff(minute,timein,timeout) as minutesClockedIn, 
        datediff(minute,timein,timeout) - @shiftMinutes as totalMinutesOffset,
        excessOrShort =
            case 
                when (datediff(minute,timein,timeout) - @shiftMinutes) > 0 then 'excess'
                when (datediff(minute,timein,timeout) - @shiftMinutes) < 0 then 'short'
                else 'ontime'
            end
    from @attendance
)
-- use the CTE
select Date,
    minutesClockedIn, 
    totalMinutesOffset,
    totalMinutesOffset / 60 as hoursOffset,
    totalMinutesOffset % 60 as minutesOffset,
    excessOrShort,
    excessOrShortDescription =
        case 
            when excessOrShort in ('excess','short')
            then right('0' + cast(abs(totalMinutesOffset / 60) as varchar(2)),2) + ':' + right('0' + cast(abs(totalMinutesOffset % 60) as varchar(2)),2) + ' ' + excessOrShort
            else 'ontime'
        end
from shiftData
于 2013-06-06T00:50:34.210 回答