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Microsoft Access 2013 Report Grouping levels need to count unique records

Posted on 2014-11-14
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Last Modified: 2014-11-18
I have a report almost complete. I track request from various sources. The database tracks the time spent on each request. dbreqid is the key field and the request has multiple entries for documenting time. When I do a group by requester and count the number of request it counts for each of my time records. One request will have multiple time records. How do I count just the distinct records in the dbreqid field. I have group level for the source but it counts each individual time record not each dbreqid. I tried to group on dbreqid but that doesn't work either. Thanks for the help.
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Question by:marlind605
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Gustav Brock earned 500 total points
ID: 40442753
You'll have to use a subquery, something like this:

Select
    T.Source,
    T.Requester,
    T.TotalTime,
    Count(*) As Requests
From
    (Select Source, Requester, Sum(TimeSpent) As TotalTime From tblTable Group By Source, Requester) As T
Group By
    T.Source,
    T.Requester,
    T.TotalTime

This should give you the clean figures.
In your report, however, you would do the grouping and counting, Thus it should use the subquery above as source.

/gustav
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Expert Comment

by:PatHartman
ID: 40442759
You may need to break the report into a main report with a subreport.  That will allow you to count records separately for each recordset.
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Author Comment

by:marlind605
ID: 40443585
I am looking at the two answers. Thanks.
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Author Comment

by:marlind605
ID: 40448803
Still not getting this. Tried subreports and different queries. Any other help?
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Expert Comment

by:Gustav Brock
ID: 40449329
That's hard without knowing what you did try.

/gustav
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Author Closing Comment

by:marlind605
ID: 40449899
What I ended up doing was creating two make table queries one to show each request by organization and the other created from the first result to show count by organization and request and then used the added the query to the one I was using for the report. It is not the way I wanted to go but seems to run quick and is accurate. There may be another solution but I couldn't figure it out. Thanks for all the replies.
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Expert Comment

by:Gustav Brock
ID: 40449938
OK, thanks for the feedback.

/gustav
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