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Running Count of records per emloyees in access query

Posted on 2007-03-23
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Last Modified: 2013-11-27
I am creating an access database query and I need to add a field that will have a running count for each employee.

This is what I have
Gary  1-5-07
Gary  1-12-07
Gary  1-19-07
Jeff    1-12-07
Jeff    1-19-07
Bob   1-5-07
Bob   1-12-07
Bob   1-19-07

and this is what I need:

Gary  1-5-07    1
Gary  1-12-07  2
Gary  1-19-07  3
Jeff    1-12-07  1
Jeff    1-19-07  2
Bob   1-5-07    1
Bob   1-12-07  2
Bob   1-19-07  3

I can't figure out what to do here...

Thanks a lot,
Bonnie
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Question by:Bonnie_K
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8 Comments
 
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Expert Comment

by:jerryb30
ID: 18783297
Would any user have two entries for the same date?
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Assisted Solution

by:GRayL
GRayL earned 800 total points
ID: 18783304
SELECT a.Name, a.MyDate, (Select Count(b.Name) FROM myTable AS b WHERE b.Name=a.Name and b.MyDate<=a.MyDate) as Ctr FROM myTable AS a;
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Author Comment

by:Bonnie_K
ID: 18783386
Jerryb,

No, the dates are distinct per employee.

I am going to try GrayL's idea now
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Accepted Solution

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dbeneit earned 1200 total points
ID: 18786388
GRayL solution is ok, but with big tables, are slow because he use fields of external query into subquery.

If yours tables are bigs try this
SELECT a.Name, a.MyDate, count(b.Name) from FROM myTable AS a inner join myTable AS b on
a.Name = b.Name and b.MyDate<=a.MyDate group by a.Name, a.MyDate;
It is the same but converting to join isntead subquery.
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Expert Comment

by:GRayL
ID: 18786822
Do you mean 'Group By' or 'Order By'?
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Expert Comment

by:dbeneit
ID: 18786906
Grayl,
group by , becouse we need only 1 row for each row of table a.
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Expert Comment

by:Jeffrey Coachman
ID: 18790910
Bonnie_K:,

Why not do this in a Report?

Report TextBoxes have a "Running Total" property, that you can set to Over Group or Over All.

So if you grouped your Report by Name you would select "Over Group"
https://filedb.experts-exchange.com/incoming/ee-stuff/2979-Access--Ruuning-GroupTotals.zip

Is this what you wanted?

Jeff Coachman
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Author Comment

by:Bonnie_K
ID: 18793345
Thank you very much,

Bonnie
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