SHOW OFF DAY

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SHOW OFF DAY

immaduddinahmed

Ten Centuries

Points: 1375
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June 15, 2013 at 3:16 am

#273981

HI
i want to show off days in my report
my sql data is like this
date----------timein------timeout--shift----------eid
2-May-2013--9:20AM---6:17PM-----G-----------17090
3-May-2013--9:09AM---2:01PM-----G-----------17090
4-May-2013--9:03AM---10:41AM----G-----------17090
sql data doesnot show 5 may data in sql or u can say sunday data becouse sunday is off and no employee come on sunday thats why i cant show off day in report
please give me a suggestion to look data like this
or any month when sunday come report show O in shift and show date just like below
date----------timein------timeout--shift----------------eid
2-May-2013--9:20AM---6:17PM-----G----------------17090
3-May-2013--9:09AM---2:01PM-----G----------------17090
4-May-2013--9:03AM---10:41AM----G----------------17090
5-May-2013--------------------------O----------------17090
i hope u under stand
Thank in advance
immad

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immaduddinahmed Ten Centuries Points: 1375 More actions · Answer 1

This is my calendar data structure

CREATE TABLE [dbo].[Calendar]

(

[Date] [datetime] NULL

)

i insert all 2013 months dates in this table

i want to show offday dates in database like this

date----------------------------eid----shift--------timein---------------------timeout--

2013-05-04 00:00:00.000--17031----G----2013-06-13 09:15:00.000-----2013-06-13 15:23:00.000

2013-05-04 00:00:00.000--17031----O-----------NULL------------------------NULL

please help me out

immad

Dird SSCertifiable Points: 5081 More actions · Answer 2

select c.DATE, s.eid, isnull(s.Shift,'O'), s.timein, s.timeout

from Shift s right outer join calendar c on s.date=c.date

Like this? The "shift" table would be replaced with the table containing "timein" etc

Dird

immaduddinahmed Ten Centuries Points: 1375 More actions · Answer 3

sir please implemenet on this query

select

[date],

min([Timein]) as First_Record,

sum(DATEDIFF(minute, [Timein], [Timeout])) as Time_Minutes

into #temp1 from Atend

where eid = 26446

group by [date]

GO

select

t.[date],

t.eid,

t.[Timein] as timein,

t.[Timeout] as timeout,

CONVERT(VARCHAR(8), DATEADD(minute, Time_Minutes, 0), 108) AS SpendTime,

case when (540 - Time_Minutes) > 0 Then '- ' else '+ ' end

+CONVERT(VARCHAR(8), DATEADD(minute, ABS(540 - Time_Minutes), 0), 108) as excesshorttime

FROM Atend t

left join #temp1 t2 on t.[date] = t2.[date] and t.[Timein] = t2.First_Record

where eid = 26446

order by t.[date], t.[Timein]

i want this type of data

date----------------------------eid----shift--------timein---------------------timeout--

2013-05-04 00:00:00.000--17031----G----2013-06-13 09:15:00.000-----2013-06-13 15:23:00.000

2013-05-04 00:00:00.000--17031----O-----------NULL------------------------NULL

immad

Dird SSCertifiable Points: 5081 More actions · Answer 4

Why would "2013-05-04" be an off day AND 'G'?

Why does your query have SpendTime, excesshorttime when the example of what you want doesn't have it?

You also give no info about the data -_ I thought eid = employee id but then it doesn't make sense why you look for the minimum timein (since an employee would only timein once per day?). You also have no "shift" column so I assumed you want a case. The solution below may work although it won't be the best.

with t1 as (select [date], min([Timein]) as First_Record,

sum(DATEDIFF(minute, [Timein], [Timeout])) as Time_Minutes

from Atend where eid = 26446

group by [date]),

t2 as (select

t.[date],t.eid,t.[Timein] as timein,t.[Timeout] as timeout,

CONVERT(VARCHAR(8), DATEADD(minute, Time_Minutes, 0), 108) AS SpendTime,

case when (540 - Time_Minutes) > 0 Then '- ' else '+ ' end

+CONVERT(VARCHAR(8), DATEADD(minute, ABS(540 - Time_Minutes), 0), 108) as excesshorttime

FROM Atend t inner join t1 t1 on t.[date] = t1.[date] and t.[Timein] = t1.First_Record

where eid = 26446)

select c.date, isnull(t2.eid,26446) eid,

case when t2.timein is null then 'O'

else 'G' end as Shift, t2.timein, t2.timeout

from calendar c left outer join t2 t2 on c.date= t2.date

order by c.[date], t2.[timein]

Dird

Jeff Moden SSC Guru Points: 1003851 More actions · Answer 5

I'd like to recommend that you should not have separate date and time columns for this because it will someday become a huge pain when someone works through midnight.

--Jeff Moden

RBAR is pronounced "ree-bar" and is a "Modenism" for Row-By-Agonizing-Row.
First step towards the paradigm shift of writing Set Based code:
________Stop thinking about what you want to do to a ROW... think, instead, of what you want to do to a COLUMN.

Change is inevitable... Change for the better is not.

Helpful Links:
How to post code problems
How to Post Performance Problems
Create a Tally Function (fnTally)

immaduddinahmed Ten Centuries Points: 1375 More actions · Answer 6

i am making attendance project for a factory.employee come in factory time in and time out. on sunday employee have off day and some time employee get absent in my data i dont have absent and off day data i want to create this. my data is like this

date---------------------------eid------------------timein-----------------------------timeout-------------------------------spendtime---------excesshsort

2013-05-02 00:00:00.000--17031----2013-06-13 09:34:00.000---2013-06-13 18:30:00.000-------08:56:00------ 00:04:00

2013-05-03 00:00:00.000--17031------2013-06-13 09:55:00.000------2013-06-13 13:30:00.000-------04:41:00--------- 04:19:00

i want this type of data

date---------------------------eid----------------timein----------------------------------timeout------------------------spendtime---------excesshsort

2013-05-02 00:00:00.000--17031-----2013-06-13 09:34:00.000------2013-06-13 18:30:00.000-------08:56:00------ 00:04:00

2013-05-03 00:00:00.000--17031------2013-06-13 09:55:00.000------2013-06-13 13:30:00.000-------04:41:00------ 04:19:00

2013-05-04 00:00:00.000---17031---------2013-06-13 09:15:00.000------2013-06-13 15:23:00.000-----06:08:00-- 02:52:00

2013-05-05 00:00:00.000--17031--------------NULL------------------------NULL-------------------------------NULL---------NULL

when employee absent or off day data show in table off day mean sunday

i hope u under stand

thanks

immad

ChrisM@Work SSC Guru Points: 186127 More actions · Answer 7

There's another element to this issue here. The OP has a calendar table but is unsure how to plug it into the query above.

^{“Write the query the simplest way. If through testing it becomes clear that the performance is inadequate, consider alternative query forms.” - Gail Shaw}

For fast, accurate and documented assistance in answering your questions, please read this article.
Understanding and using APPLY, (I) and (II) Paul White
Hidden RBAR: Triangular Joins / The "Numbers" or "Tally" Table: What it is and how it replaces a loop Jeff Moden

aaron.reese SSChampion Points: 13494 More actions · Answer 8

The answer to the OP question is to use LEFT JOIN

SELECT Calendar.date, shiftQuery.* From Calendar LEFT JOIN shiftQuery on Calendar.date = shiftQuery.date

This will give you the calendar record EVERY TIME and the shift record IF IT EXISTS

Dird SSCertifiable Points: 5081 More actions · Answer 9

I'm guessing it would actually be used in an OUTER JOIN like in the possible solution I gave~ from what he says the main table doesn't contain saturday/sunday records so he created the calendar for it.

@OP: did you try the code I sent back?

Dird