需要Sql选择查询员工考勤月报

我有三张桌子 1.日历日期 2.员工详细信息 3.打孔计时 根据下表，我需要生成月度出勤报告

date
2017-06-01
2017-06-02
2017-06-03
2017-06-04
2017-06-05
2017-06-06
2017-06-07
2017-06-08
2017-06-09
2017-06-10
2017-06-11
2017-06-12
2017-06-13
2017-06-14
2017-06-15


pk  EmployeeName
3   IMMAM
4   SRIMAN
5   AJAY
6   Vinay
7   RAGHU

三,。打孔计时

EmployeePk  PunchTime
4   2017-06-10 17:49:34.000
4   2017-06-10 18:51:35.000
7   2017-06-10 19:18:37.000
6   2017-06-10 19:18:52.000
5   2017-06-10 19:19:05.000
6   2017-06-12 10:59:34.000
5   2017-06-12 10:59:44.000
5   2017-06-12 16:04:24.000
6   2017-06-12 16:06:48.000
5   2017-06-12 16:08:58.000
5   2017-06-12 16:14:33.000
5   2017-06-13 10:44:06.000
6   2017-06-13 10:44:23.000
4   2017-06-13 10:44:36.000
7   2017-06-13 10:51:22.000
5   2017-06-13 17:45:59.000
6   2017-06-13 17:46:14.000
7   2017-06-13 17:46:26.000
4   2017-06-13 17:47:21.000
5   2017-06-14 10:48:39.000
6   2017-06-14 10:49:04.000
7   2017-06-14 10:49:16.000
4   2017-06-14 10:49:23.000
6   2017-06-14 17:22:34.000
7   2017-06-14 18:23:08.000
4   2017-06-14 18:23:25.000
5   2017-06-14 18:23:32.000
5   2017-06-15 10:44:48.000
4   2017-06-15 10:45:13.000
6   2017-06-15 10:45:32.000
7   2017-06-15 11:03:55.000
5   2017-06-15 11:26:53.000
7   2017-06-15 11:26:56.000
5   2017-06-15 11:29:16.000
5   2017-06-15 11:29:20.000

我需要这样的报告

Calender Date   Employee    Punchtime   IsPresent
10-06-2017  Imama   Null    No
10-06-2017  SRIMAN  2017-06-10 17:49:34.000 Yes
10-06-2017  AJAY    2017-06-10 19:19:05.000 Yes
10-06-2017  Vinay   2017-06-10 19:18:52.000 Yes
10-06-2017  RAGHU   2017-06-10 19:18:37.000 Yes
11-06-2017  Imama   Null    NO
11-06-2017  SRIMAN  Null    NO
11-06-2017  AJAY    Null    NO
11-06-2017  Vinay   Null    NO
11-06-2017  RAGHU   Null    NO
12-06-2017  Imama   Null    No
12-06-2017  SRIMAN  Null    No
12-06-2017  AJAY    2017-06-12 10:59:44.000 Yes
12-06-2017  Vinay   2017-06-12 10:59:34.000 Yes
12-06-2017  RAGHU   2017-06-12 10:59:44.000 Yes

还有秀安

我是这样开始的

declare
@month int, @year int, @lastDay int,
@starttime datetime,
@endtime datetime,
@companyPk int,
@employeePk int

set @starttime ='2017-06-01'
set @endtime = '2017-06-30'
set @companyPk =2
set @employeePk =0

select x.calenderdate,y.Pk,y.EmployeeName,y.Ispresent 

from
 (

(select convert(date, CalendarDate) as calenderdate from ECMSCalendar
 where convert(date, CalendarDate) >= @starttime and convert(date, CalendarDate) <=@endtime
) as x
left join 
(
select a.Pk,a.EmployeeName,(case when b.IsPresent is null then 'Absent' else b.IsPresent end) as Ispresent  from 
(
(select emp.Pk,emp.EmployeeName from EmployeeDetails emp where emp.Companypk = @companyPk) as  a
left join 
(SELECT ed.pk as Pk,ed.EmployeeName as StaffName , 

(case when (DATEDIFF(MINUTE,min(bio.PunchTime),(case when max(bio.PunchTime) = min(bio.PunchTime) then min(bio.PunchTime) else max(bio.PunchTime) end) )) < sd.shiftname then 'Absent' else 'Present' end) as IsPresent

--, CONVERT(nvarchar(2), bio.PunchTime) as date
--,count(CONVERT(nvarchar(2), bio.punchtime, 103) )as day

from BioMetricPunchDetails bio
left join EmployeeDetails ed on ed.Pk = bio.EmployeePk
left join EmployeeShiftDetails esd on esd.EmployeePk = ed.pk
left join ShiftDetails sd on sd.pk  =esd.shiftpk
where --convert (date,bio.punchtime) >= @starttime and convert (date,bio.punchtime) <= @endtime
convert (date,(select min(bpd.PunchTime) from BioMetricPunchDetails bpd where (bpd.EmployeePk=@employeePk or @employeePk=0))) >= @starttime and convert (date,(select max(bpd.PunchTime) from BioMetricPunchDetails bpd where (bpd.EmployeePk=@employeePk or @employeePk=0))) <= @endtime
and 
ed.Companypk= @companyPk and (ed.Pk=@employeePk or @employeePk=0) 
--and ed.UpdatedByUserPk=@days 
group by ed.pk, ed.EmployeeName,sd.shiftname) as b 
on a.Pk = b.Pk)
) as y 
on x.calenderdate = y.Ispresent
)
order by x.calenderdate

---

请解决这个问题谢谢你试试这样的方法

with 
alldates as (
SELECT convert(date,thedate) alldate
FROM dbo.ExplodeDates( '2017-06-10' , '2017-06-10') as d
),

empshift as (
select ed.Pk,ed.EmployeeName,sd.ShiftName,sd.Intime,sd.OutTime from EmployeeDetails ed left outer join 
EmployeeShiftDetails esd on (ed.Pk = esd.EmployeePk)
left outer join ShiftDetails sd on (sd.Pk = esd.ShiftPk ) 
)
,biometric1 as (
select min(punchtime) intime,max(punchtime) outtime,convert(date,punchtime) dates

, empshift.EmployeeName,empshift.ShiftName,Intime ShiftInTime,OutTime ShiftOutTime
from BioMetricPunchDetails bpd inner join empshift on (empshift.Pk =bpd.EmployeePk )

where CONVERT(date,punchtime) between '2017-06-10' and '2017-06-0'

group by convert(date,punchtime), empshift.EmployeeName,empshift.ShiftName,Intime,OutTime
),

fnl as (
select convert(date,intime) dates,intime,
case when (intime = outtime ) then NULL else outtime end as outtime
,ShiftName,DATEDIFF(MINUTE,intime,outtime) duration,ShiftInTime,convert(time,intime) InTimeHrMin
,datediff(MINUTE,ShiftInTime, convert(time,intime)) lateby
 from biometric1
 ),

fnl1 as (select * from alldates left outer join fnl
on (alldates.alldate = fnl.dates)) 

select ed.EmployeeName,fnl1.* from EmployeeDetails ed cross join fnl1 

报告很好。他们会发送整洁的小电子邮件或PowerPoint 没有人真正关注的演讲，但它们确实让我们 感觉我们很有效率

1.您是否在跟踪哪个deadbeat错过了月刊 会议

你的会议怎么了？他们是无聊的节日吗？他们真的提供了团队间的凝聚力吗，就像远足或IM一样 Skype、微软团队、Slack无法提供？ 2.你是否想看看谁在忙，谁来参加会议

打卡至少表明他们在那里……但他们有吸引力吗？增加价值？ 当然，还有其他指标可以帮助你发现同事的职业道德。 让您的报告回答简单查询无法回答的问题。避免报告的诱惑，因为它们会填满空间

无论您的公司如何跟踪会议，请务必以表格形式获取日期

CREATE TABLE #Calendar_Date (DATES Date)
INSERT INTO #Calendar_Date
VALUES ('2017-06-01'),('2017-06-02'),('2017-06-03'),('2017-06-04'),('2017-06-05'),('2017-06-06'),('2017-06-07'),('2017-06-08'),('2017-06-09'),('2017-06-10'),('2017-06-11'),('2017-06-12'),('2017-06-13'),('2017-06-14'),('2017-06-15')

CREATE TABLE #Employee_DIM (ID INT UNIQUE
                         , Employee_Name NVARCHAR(100) )
INSERT INTO #Employee_DIM (ID, Employee_Name)
VALUES (3, 'IMMAM'), (4, 'SRIMAN'), (5, 'AJAY'), (6, 'Vinay'), (7, 'RAGHU')

CREATE TABLE #Punch_Timings (Employee_PK INT
                          , Punch_Time DATETIME)

INSERT INTO #Punch_Timings (Employee_PK, Punch_Time)
VALUES (4,'2017-06-10 17:49:34.000'),(4,'2017-06-10 18:51:35.000'),(7,'2017-06-10 19:18:37.000')
      ,(6,'2017-06-10 19:18:52.000'),(5,'2017-06-10 19:19:05.000'),(6,'2017-06-12 10:59:34.000')
      ,(5,'2017-06-12 10:59:44.000'),(5,'2017-06-12 16:04:24.000'),(6,'2017-06-12 16:06:48.000')
      ,(5,'2017-06-12 16:08:58.000'),(5,'2017-06-12 16:14:33.000'),(5,'2017-06-13 10:44:06.000')
      ,(6,'2017-06-13 10:44:23.000'),(4,'2017-06-13 10:44:36.000'),(7,'2017-06-13 10:51:22.000')
      ,(5,'2017-06-13 17:45:59.000'),(6,'2017-06-13 17:46:14.000'),(7,'2017-06-13 17:46:26.000')
      ,(4,'2017-06-13 17:47:21.000'),(5,'2017-06-14 10:48:39.000'),(6,'2017-06-14 10:49:04.000')
      ,(7,'2017-06-14 10:49:16.000'),(4,'2017-06-14 10:49:23.000'),(6,'2017-06-14 17:22:34.000')
      ,(7,'2017-06-14 18:23:08.000'),(4,'2017-06-14 18:23:25.000'),(5,'2017-06-14 18:23:32.000')
      ,(4,'2017-06-15 10:45:13.000'),(6,'2017-06-15 10:45:32.000'),(7,'2017-06-15 11:03:55.000')
      ,(5,'2017-06-15 11:26:53.000'),(7,'2017-06-15 11:26:56.000'),(5,'2017-06-15 11:29:16.000')
      ,(5,'2017-06-15 11:29:20.000'),(4, '2017-06-10 19:31:35.000'),(5, '2017-06-10 19:45:05.000')
      ,(6, '2017-06-10 19:38:52.000'),(7, '2017-06-10 19:13:37.000')

CREATE TABLE #MEETING_TIMES (Meeting_ID INT
                           , Meeting_Start DATETIME
                           , Meeting_End DATETIME)
INSERT INTO #MEETING_TIMES (Meeting_ID, Meeting_Start, Meeting_End)
VALUES (304,'2017-06-10 19:00:00.000', '2017-06-10 19:30:00.000')
     , (311, '2017-06-14 18:00:00.000', '2017-06-14 18:30:00.000')

一旦你有了有序的维度和事实，逻辑应该非常简单，如下所示：

;WITH Employee_Time AS (
SELECT MIN(Punch_Time) AS ASSUME_Punch_IN
     , MAX(PUNCH_Time) AS ASSUME_Punch_OUT
     , DATEPART(DAY, Punch_Time) AS DAY
     , Employee_PK
FROM #Punch_Timings PT
GROUP BY Employee_PK, DATEPART(DAY, Punch_Time) 
)

SELECT B.Meeting_ID
     , A.Meeting_Start
     , Employee_ID
     , CASE WHEN A.Meeting_Start IS NULL THEN 'NO' ELSE 'YES' END AS FLAG_ATTEND
FROM (SELECT ET.Employee_PK, ET.ASSUME_Punch_IN
           , ET.ASSUME_Punch_OUT, ET.DAY, MT.Meeting_Start, MT.Meeting_End
      FROM Employee_Time ET
      INNER JOIN #MEETING_TIMES MT ON ( (   ET.ASSUME_Punch_IN >= MT.Meeting_Start
                                        AND ET.ASSUME_Punch_IN < MT.Meeting_End)
                                    OR (    ET.ASSUME_Punch_IN <= Meeting_Start
                                        AND ET.ASSUME_Punch_OUT > Meeting_Start)
                                     )
                                 AND ET.DAY = DATEPART(DAY, Meeting_Start) 
            )  A
RiGHT OUTER JOIN ( 
                    SELECT ID AS Employee_ID, Meeting_Start, Meeting_ID
                   FROM #MEETING_TIMES  B       
                   FULL JOIN #Employee_DIM ED ON 1=1
                   ) B ON B.Employee_ID = A.Employee_PK
                      AND DATEPART(DAY, B.Meeting_Start) = A.DAY

---

也就是说，我会从中找到其他信息，比如你同事项目的紧迫性和性质。在数据中找到一个不容易回答的问题，并让您的报告回答这个问题。

太好了！您应该从SELECT开始。查看LEFT JOIN的作用。似乎太复杂了…格式化查询也会有很大帮助。您的数据不干净，因此假定打孔是有问题的。你需要回答的问题是什么？谢谢你，我会尽力回去的。顺便说一句，第一个问题的答案是3：用于将会议与打卡时间匹配的逻辑是IF Punch\u IN>=会议开始，IF Punch\u IN<会议结束，否则IF Punch\u IN会议开始