如何分组时间段并检查休息时间
我有一个存储功能，可以提取所有员工的时钟信息。我正试图提出一份例外报告来审核午餐。我当前的查询一次构建所有信息1段。
SELECT ftc.lEmployeeID, ftc.sFirstName, ftc.sLastName, ftc.dtTimeIn,
ftc.dtTimeOut, ftc.TotalHours, ftc.PunchedIn, ftc.Edited
FROM dbo.fTimeCard(@StartDate, @EndDate, @DeptList,
@iActive, @EmployeeList) AS ftc
LEFT OUTER JOIN Employees AS e ON ftc.lEmployeeID = e.lEmployeeID
WHERE (ftc.TotalHours >= 0) AND (ftc.DID IS NOT NULL) OR
(ftc.DID IS NOT NULL) AND (ftc.dtTimeOut IS NULL)
此输出如下所示：
24 Bob bibby 8/2/2013 11:55:23 AM 8/2/2013 3:36:44 PM 3.68
24 bob bibby 8/2/2013 4:10:46 PM 8/2/2013 8:14:30 PM 4.07
39 rob blah 8/2/2013 8:01:57 AM 8/2/2013 5:01:40 PM 9.01
41 john doe 8/2/2013 10:09:58 AM 8/2/2013 1:33:38 PM 3.4
41 john doe 8/2/2013 1:55:56 PM 8/2/2013 6:10:15 PM 4.25
我需要查询才能做两件事。
1）每天将这些细分组合在一起。 2）在新列中报告“休息时间”
获得该信息后，我需要检查每个细分的小时数，并确保发生两件事。
1）如果他们共工作了6个小时，他们会休息30分钟吗？ 2）如果他们休息一下，他们会休息一下吗？ 30分钟。
你看到鲍勃在上午11点55分打了一拳，并在3点36分打了一顿午餐。他于4点10分午餐时冲了上来，并于8点14分拳打脚踢。他共工作了7.75小时，并且休息了34分钟。 他在这里很好。我不想报告例外
约翰总共工作了7.65个小时。然而，当他拳打脚踢时，他只吃了22分钟的午餐。我需要报告“吉姆只吃了22分钟的午餐”你也会看到抢劫工作了9个小时，没有休息。我需要报告“抢劫工作超过6小时并且没有休息”
我认为如果我可以完成对两个细分的分组。然后我可以处理报告方面。
的 的 *更新**
我更改了查询以尝试完成此操作。以下是我目前的疑问：
SELECT ftc.lEmployeeID, ftc.sFirstName, ftc.sLastName, ftc.TotalHours, DATEDIFF(mi, MIN(ftc.dtTimeOut), MAX(ftc.dtTimeIn)) AS Break_Time_Minutes
FROM dbo.fTimeCard(@StartDate, @EndDate, @DeptList, @iActive, @EmployeeList) AS ftc LEFT OUTER JOIN
Employees AS e ON ftc.lEmployeeID = e.lEmployeeID
WHERE (ftc.TotalHours >= 0) AND (ftc.DID IS NOT NULL) OR
(ftc.DID IS NOT NULL) AND (ftc.dtTimeOut IS NULL)
GROUP BY ftc.lEmployeeID, ftc.sFirstName, ftc.sLastName, ftc.TotalHours
我的输出目前看起来像这样：
24 Bob bibby 3.68 -221
24 bob bibby 4.07 -244
39 rob blah 0.05 -3
39 rob blah 2.63 -158
41 john doe 3.4 -204
41 john doe 4.25 -255
正如您所看到的那样，它并没有按日期组合细分，而Break_time显示的是负分钟。它也没有结合日子。鲍勃的时间应该在1行。并且显示7.75分钟的休息时间应该是34分钟。
7 个答案:
答案 0 :(得分：4)
我相信如果你想要将两个时间结合起来，你需要将它们从组中取出并加上它们。根据结果，报告可以检查总小时数和休息时间。如果要标记它们，可以添加case语句。
SELECT ftc.lEmployeeID
,ftc.sFirstName
,ftc.sLastName
,SUM(ftc.TotalHours) AS TotalHours
,DATEDIFF(mi, MIN(ftc.dtTimeOut), MAX(ftc.dtTimeIn)) AS BreakTimeMinutes
FROM dbo.fTimeCard(@StartDate, @EndDate,
@DeptList, @iActive,@ EmployeeList) AS ftc
WHERE SUM(ftc.TotalHours) >= 0 AND (ftc.DID IS NOT NULL) OR
(ftc.DID IS NOT NULL) AND (ftc.dtTimeOut IS NULL)
GROUP BY ftc.lEmployeeID, ftc.sFirstName, ftc.sLastName
我在sql中进行了这个快速测试，它似乎按照你想要的方式工作。你有没有给小组添加一些东西？
declare @table table (emp_id int,name varchar(4), tin time,tout time);
insert into @table
VALUES (1,'d','8:30:00','11:35:00'),
(1,'d','13:00:00','17:00:00');
SELECT t.emp_id
,t.name
,SUM(DATEDIFF(mi, tin,tout))/60 as hours
,DATEDIFF(mi, MIN(tout), MAX(tin)) AS BreakTimeMinutes
FROM @table t
GROUP BY t.emp_id, t.name
答案 1 :(得分：2)
使用示例SQL的相关部分，我创建了一个SQL Fiddle，显示了如何完成此操作。您可以在此处查看：http://sqlfiddle.com/#!6/f05ce/3
SELECT EmployeeId, Num_Hours,
CASE WHEN tmp.Break_Time_Minutes < 0 Then 0 Else Break_Time_Minutes END As Break_Time_Minutes,
CASE WHEN tmp.Break_Time_Minutes < 0 Then 1 Else 0 END As SkippedBreak
FROM (
SELECT EmployeeId,
Round(SUM(DATEDIFF(second, TimeIn, TimeOut) / 60.0 / 60.0),1) As NUM_Hours,
DateDiff(mi, Min(TimeOut), Max(TimeIn)) As Break_Time_Minutes FROM Employee
GROUP BY EmployeeId, CAST(TimeIn As Date)
) as tmp WHERE tmp.Num_Hours > 6 AND Break_Time_Minutes < 30
答案 2 :(得分：2)
自包含示例：
这是一个如何工作的例子。
我使用的独立逻辑不依赖于您自定义的功能，因为此处的社区无法访问它（包括它使用的数据）。
相反，我建立了你设法提供的“输出”的答案
这将自包含，因为没有对象依赖。
DECLARE @EmpClock Table
(
EmployeeID Int,
FirstName VarChar(50),
LastName VarChar(50),
PunchIn DateTime,
PunchOut DateTime
)
INSERT INTO @EmpClock
SELECT 24,'Bob','bibby','8/2/2013 11:55:23 AM','8/2/2013 3:36:44 PM' UNION
SELECT 24,'bob','bibby','8/2/2013 4:10:46 PM','8/2/2013 8:14:30 PM' UNION
SELECT 39,'rob','blah','8/2/2013 8:01:57 AM','8/2/2013 5:01:40 PM' UNION
SELECT 41,'john','doe','8/2/2013 10:09:58 AM','8/2/2013 1:33:38 PM' UNION
SELECT 41,'john','doe','8/2/2013 1:55:56 PM','8/2/2013 6:10:15 PM' UNION
SELECT 1,'Mike','TeeVee','8/2/2013 12:05:30 PM','8/2/2013 2:15:45 PM' UNION
SELECT 1,'Mike','TeeVee','8/2/2013 2:25:05 PM','8/2/2013 3:35:25 PM' UNION
SELECT 1,'Mike','TeeVee','8/2/2013 3:50:15 PM','8/2/2013 5:30:55 PM' UNION
SELECT 1,'Mike','TeeVee','8/2/2013 5:40:35 PM','8/2/2013 6:50:20 PM'
SELECT *,
DATEDIFF(SECOND, '', EC.WorkedTotal)/60.0/60.0[WorkedHours],
DATEDIFF(SECOND, '', EC.BreakTotal )/60.0 [BreakMinutes]
FROM
(
SELECT EC.*,
DATEADD(SECOND, DATEDIFF(SECOND, WorkedTotal, WorkPeriod), CAST('' as Time(0)))[BreakTotal]
FROM
(
SELECT EC.EmployeeID, EC.EmployeeName, EC.Day,
DATEADD(SECOND, DATEDIFF(SECOND, EC.FirstPunchIn, EC.LastPunchOut), CAST('' as Time(0)))[WorkPeriod],
DATEADD(SECOND, EC.Worked, CAST('' as Time(0)))[WorkedTotal]
FROM
(
SELECT EC.EmployeeID,
(EC.FirstName + ' ' + EC.LastName)[EmployeeName],
--"Day" Assumes Punches do not span across midnight.
-- If any do, then the day of the Punch-In will be used.
CAST(EC.PunchIn as Date)[Day],
SUM(DATEDIFF(SECOND, EC.PunchIn, EC.PunchOut))[Worked],
MIN(EC.PunchIn)[FirstPunchIn],
MAX(EC.PunchOut)[LastPunchOut]
FROM @EmpClock as EC
GROUP BY EC.EmployeeID, (EC.FirstName + ' ' + EC.LastName), CAST(EC.PunchIn as Date)
) AS EC
) AS EC
) AS EC
WHERE EC.BreakTotal > DATEADD(MINUTE, 30, CAST('' as Time(0)))
AND EC.WorkedTotal > DATEADD(HOUR, 6, CAST('' as Time(0)))
结果：
注意我为名为“ Mike TeeVee ”的虚构员工添加了更多数据。
我是这样做的，以防你有一个员工休息休息或因任何原因需要紧急休息
这允许我们测试逻辑如何处理这些情况。
如果没有Where-Clause，我们会看到：
按原样运行（使用Where-Clause），您将看到它正确过滤掉结果：
您会注意到我将结果的Time-DataType格式显示为“ WorkedTotal ”和“ BreakTotal ”。我更喜欢这个用于显示目的，因为我们通常不会将小时视为整体100％的分数，而是将小时和剩余分数视为分钟。
我继续将分数小时和分数分钟分别包括为“ WorkedHours ”和“ BreakMinutes ”，以防您的要求需要以此格式进一步计算
“ WorkPeriod ”表示班次的持续时间（包括休息时间）。
我确定你不需要这些信息，但我把它包括在内是为了完整。
使用您自己的逻辑：
从这里的其他答案来看，我发现你在将你的逻辑融入他们的答案时遇到了问题，所以我也为你做了这个。
以下脚本只能在您的环境中运行：
SELECT *,
DATEDIFF(SECOND, '', EC.WorkedTotal)/60.0/60.0[WorkedHours],
DATEDIFF(SECOND, '', EC.BreakTotal )/60.0 [BreakMinutes]
FROM
(
SELECT EC.*,
DATEADD(SECOND, DATEDIFF(SECOND, WorkedTotal, WorkPeriod), CAST('' as Time(0)))[BreakTotal]
FROM
(
SELECT EC.EmployeeID, EC.EmployeeName, EC.Day,
DATEADD(SECOND, DATEDIFF(SECOND, EC.FirstPunchIn, EC.LastPunchOut), CAST('' as Time(0)))[WorkPeriod],
DATEADD(SECOND, EC.Worked, CAST('' as Time(0)))[WorkedTotal]
FROM
(
SELECT EC.EmployeeID,
(EC.FirstName + ' ' + EC.LastName)[EmployeeName],
--"Day" Assumes Punches do not span across midnight.
CAST(EC.PunchIn as Date)[Day],
SUM(DATEDIFF(SECOND, EC.PunchIn, EC.PunchOut))[Worked],
MIN(EC.PunchIn)[FirstPunchIn],
MAX(EC.PunchOut)[LastPunchOut]
FROM
( --I replaced my table variable @EmpClock with a call to your own logic.
SELECT EC.lEmployeeID[EmployeeID], EC.sFirstName[FirstName], EC.sLastName[LastName],
EC.dtTimeIn[PunchIn], EC.dtTimeOut[PunchOut]
--You have these in your original query, but the values are missing in your "output".
--,EC.TotalHours, EC.PunchedIn, EC.Edited
FROM dbo.fTimeCard(@StartDate, @EndDate, @DeptList, @iActive, @EmployeeList) as EC
LEFT JOIN Employees as E
ON EC.lEmployeeID = E.lEmployeeID
) AS EC
GROUP BY EC.EmployeeID, (EC.FirstName + ' ' + EC.LastName), CAST(EC.PunchIn as Date)
) AS EC
) AS EC
) AS EC
WHERE EC.BreakTotal > DATEADD(MINUTE, 30, CAST('' as Time(0)))
AND EC.WorkedTotal > DATEADD(HOUR, 6, CAST('' as Time(0)))
可能的错误：
我还注意到你的where子句有问题：
WHERE (ftc.TotalHours >= 0) AND (ftc.DID IS NOT NULL) OR
(ftc.DID IS NOT NULL) AND (ftc.dtTimeOut IS NULL)
目前还不清楚你在这里尝试“ OR ”是什么，并且你有两次列出“ ftc.DID IS NOT NULL ”。
您可能希望查看该逻辑，并在使用OR时考虑正确使用括号
由于这种混淆，我在上面的例子中省略了这个逻辑。
你偶然的意思是这个吗？：
WHERE ftc.DID IS NOT NULL
AND (ftc.TotalHours >= 0 OR ftc.dtTimeOut IS NULL)
答案 3 :(得分：1)
试试这个： -
SELECT ftc.lEmployeeID,sum(ftc.TotalHours)as TotalHours,
ABS(DATEDIFF(mi, MIN(convert(datetime,ftc.dtTimeOut,9)), MAX(convert(datetime,ftc.dtTimeIn,9)))) AS Break_Time_Minutes
FROM dbo.fTimeCard(@StartDate, @EndDate, @DeptList, @iActive, @EmployeeList) AS ftc
LEFT OUTER JOIN
Employees AS e ON ftc.lEmployeeID = e.lEmployeeID
WHERE (ftc.TotalHours >= 0) AND (ftc.DID IS NOT NULL) OR
(ftc.DID IS NOT NULL) AND (ftc.dtTimeOut IS NULL)
group by ftc.EmployeeiD,DATE(ftc.dtTimeIn)
答案 4 :(得分：1)
很抱歉写在这里。内容可能很长。 你的问题仍然没有解决，但它似乎很容易，因为你强加给我们 dbo.fTimeCard（@StartDate，@ EndDate，@ DeptList，@ iActive，@ EmployeeList）。 我们不知道ftc.TotalHours是否已经是某事物的总和或它是什么。 你要做的只是在表变量中显示表结构和一些数据。你的解释是很好的enuf。 其次我想指出的是，你永远不会通过id，firstname，lastname 来做分组。 group by id绰绰有余。所以在你的情况下你需要CTE.i不能写整个查询因为某些东西是表格而且数据不清楚。
;with CTE as
(SELECT ftc.lEmployeeID, ftc.sFirstName, ftc.sLastName, ftc.dtTimeIn, ftc.dtTimeOut, ftc.TotalHours, ftc.PunchedIn, ftc.Edited
FROM dbo.fTimeCard(@StartDate, @EndDate, @DeptList, @iActive, @EmployeeList) AS ftc LEFT OUTER JOIN
Employees AS e ON ftc.lEmployeeID = e.lEmployeeID
WHERE (ftc.TotalHours >= 0) AND (ftc.DID IS NOT NULL) OR
(ftc.DID IS NOT NULL) AND (ftc.dtTimeOut IS NULL)
)
,CTE1 as
(
SELECT ftc.lEmployeeID
,SUM(ftc.TotalHours) AS TotalHours
MIN(ftc.dtTimeOut) MindtTimeOut, MAX(ftc.dtTimeIn) AS MAXdtTimeIn
FROM dbo.fTimeCard(@StartDate, @EndDate, @DeptList, @iActive, @EmployeeList) AS ftc
WHERE SUM(ftc.TotalHours) >= 0 AND (ftc.DID IS NOT NULL) OR
(ftc.DID IS NOT NULL) AND (ftc.dtTimeOut IS NULL)
GROUP BY ftc.lEmployeeID
)
,CTE2 as
(
join cte and cte1 on employeeid--well it depend
)
select * from cte2--this is just indicative.
答案 5 :(得分：1)
使用CTE对分段进行分组，您可以查询此CTE以检查中断持续时间。
在最终的CASE WHEN语句中添加其他检查：
; with
DailyRecords as
( -- add a rownumber to each entry by employee/day
select ROW_NUMBER() over (partition by ftc.lEmployeeId, cast(ftc.dtTimeIn as date)
order by ftc.lEmployeeId, ftc.dtTimeIn ) as rownum,
cast(ftc.dtTimeIn as date) as [Day],
ftc.lEmployeeID, ftc.sFirstName, ftc.sLastName, ftc.dtTimeIn, ftc.dtTimeOut, ftc.TotalHours
from fTimeCard(@StartDate, @EndDate, @DeptList, @iActive, @EmployeeList) ftc
),
DailyRequest as
( -- group 2 segments together (rownum 1 and 2), and report break time in a new column
select Segment1.lEmployeeId, Segment1.sFirstName, Segment1.sLastName,
Segment1.[Day],
coalesce(round(datediff(mi,Segment1.dtTimeIn, Segment1.dtTimeOut) / 60.0, 2),0) as Duration1,
coalesce(round(datediff(mi,Segment2.dtTimeIn, Segment2.dtTimeOut) / 60.0, 2), 0) as Duration2,
coalesce(round(datediff(mi,Segment1.dtTimeOut,Segment2.dtTimeIn),2),0) as BreakDuration
from DailyRecords Segment1
left join DailyRecords Segment2
on segment1.lEmployeeID = Segment2.lEmployeeID
and Segment1.[Day] = Segment2.[Day]
and Segment2.rownum = 2
where Segment1.rowNum= 1
)
-- make report from DailyRequest with remarks
select lEmployeeId, sFirstName, sLastName, [Day], Duration1, Duration2, BreakDuration,
case
when Duration1+Duration2 >= 6 and BreakDuration = 0 then 'No Break'
when Duration1+Duration2 >= 6 and BreakDuration < 30 then ltrim(str(coalesce(BreakDuration, 0))) +' mn break'
when Duration1+Duration2 >= 6 and BreakDuration > 35 then ltrim(str(coalesce(BreakDuration, 0))) +' mn break'
end as Remarks
from DailyRequest D
如果需要，添加LEFT OUTER JOIN Employee（在您的要求中没有使用它，我将其删除）
答案 6 :(得分：0)
确定休息时间的快捷方式，是最后一次退房和第一次办理登机手续，然后您有时间总计。它与ftc.TotalHours之间的差异是所用的休息时间。然后你可以添加一些代码来报告异常。 如果选择了多天，则下面的代码将不起作用，但是我们需要的只是一个日期变量，并将日期变量添加到GROUP BY子句中。
SELECT out.lEmployeeID, out.sFirstName, out.sLastName,
CASE WHEN (out.TotalHours > 6 AND out.Break_Time_Minutes < 30) THEN
'report exception' ELSE 0 END AS Exception_Status
FROM
( SELECT ftc.lEmployeeID, ftc.sFirstName, ftc.sLastName,
ABS(DATEDIFF(mi, MAX(ftc.dtTimeOut), MIN(ftc.dtTimeIn))) - SUM(ftc.TotalHours) * 60
AS Break_Time_Minutes, SUM(ftc.TotalHours) AS TotalHours
FROM dbo.fTimeCard(@StartDate, @EndDate, @DeptList, @iActive, @EmployeeList) AS ftc
LEFT OUTER JOIN Employees AS e ON ftc.lEmployeeID = e.lEmployeeID
WHERE (ftc.TotalHours >= 0) AND (ftc.DID IS NOT NULL) OR
(ftc.DID IS NOT NULL) AND (ftc.dtTimeOut IS NULL)
GROUP BY ftc.lEmployeeID) as out
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