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Query Min and difference

Hi There,

I have the attached database and I am working on the ActualVSsched query.  The query puts the InTime from tblClockTime and the InSched from tblEmployeeSchedule on the same view and I am trying to do two things:

1.There are several ClockTimes for each employee for the same day.  So for EG, Francoise clocks in on the 22nd May twice, once at 8:17 and once at 14:46.   I am trying to have the query show only the FIRST time any of the employees clock in for that day. In my example above, the query would only show the record that contains Francoise's clock time of 8:17.  

2. I want to make another field in the query that would subtract the InSched from the InTime but only if the InTime is greater than the InSched.  In plain english, if the employee is late, how late in minutes is that employee?

I am so new to databases that I am at a loss as to where to put this.  Trying to learn!

Any help appreciated.
EmployeeDatabaseV3.accdb
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colin911
Asked:
colin911
2 Solutions
 
santoshmotwaniCommented:
Please let me know if you need any help .
EmployeeDatabaseV3.accdb
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chaauCommented:
Not a problem.

Create a separate query called InOutTimesForTheDay.
The syntax of it must be:
SELECT tblClockTime.CSid, tblClockTime.ClockDate, Min(tblClockTime.InTime) AS InTime, Max(tblClockTime.OutTime) AS OutTime
FROM tblClockTime
GROUP BY tblClockTime.CSid, tblClockTime.ClockDate;

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Modify ActualVSsched to use this query instead of tblClockTime:

SELECT tblEmployee.CSid, tblEmployee.First, tblEmployee.Last, InOutTimesForTheDay.ClockDate, tblEmployeeSchedule.InSched, InOutTimesForTheDay.InTime, IIF( Datediff("n", tblEmployeeSchedule.InSched,InOutTimesForTheDay.InTime ) > 0 , Datediff("n", tblEmployeeSchedule.InSched,InOutTimesForTheDay.InTime ) , "On Time") AS ['Actual Start Time']
FROM (tblEmployee INNER JOIN tblEmployeeSchedule ON tblEmployee.[CSid] = tblEmployeeSchedule.[Csid]) INNER JOIN InOutTimesForTheDay ON tblEmployee.[CSid] = InOutTimesForTheDay.[CSid];

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I have modified the database and can upload it, but you will be able to do it yourself using the queries above
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colin911Author Commented:
Thank you both very much for your help.  Santos, you have the most elegant solution since it is only one query.  Chaau, I am learning so I like the see the alternate approach.
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