Count By Group
Jul 12, 2007
Hi,
I was wondering if there a method in Access to allow me to count the number of items by group?
For example, I have a set of data in a table like so:
SA
SA
SA
SP
SP
BM
I can choose Group By and the data will group them like SA, SP, BM; and I can choose Count to count the total number of records, but I would like for instance to be able to retrieve the following data:
SA: 3
SP: 2
BM: 1
Currently I'm exporting the list to Excel, and then using Subtotals. It does exactly what I want, but I'm pretty new to Access and wondered if there's a different way! :)
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Dec 3, 2004
Hey all!
I'm having issues trying to create a query that will take data from two tables and summarize it based on ID, Count(ID), Sum(Score), Sum(Xs). I don't have an issue doing the combining, just the summing and counting (actually, just the counting). I've tried playing directly with the SQL statement, but I keep getting errors.
This is what I have:
tblRoster
ID, Name, Sex, Youth
tblScores
ID, WeekNo, A1, A2, A3, A2X, A3X, B1, B2, B3, B2X, B3X
Sample Data for tblScores would look like this (scores actually are not the same, but you get the idea)
10, 1, 90, 89, 87, 3, 7, 87, 92, 99, 4, 7
10, 2, 90, 89, 87, 3, 7, 87, 92, 99, 4, 7
10, 3, 90, 89, 87, 3, 7, 87, 92, 99, 4, 7
10, 4, 90, 89, 87, 3, 7, 87, 92, 99, 4, 7
23, 1, 90, 89, 87, 3, 7, 87, 92, 99, 4, 7
23, 2, 90, 89, 87, 3, 7, 87, 92, 99, 4, 7
23, 3, 90, 89, 87, 3, 7, 87, 92, 99, 4, 7
23, 4, 90, 89, 87, 3, 7, 87, 92, 99, 4, 7
23, 5, 90, 89, 87, 3, 7, 87, 92, 99, 4, 7
67, 1, 90, 89, 87, 3, 7, 87, 92, 99, 4, 7
67, 2, 90, 89, 87, 3, 7, 87, 92, 99, 4, 7
67, 3, 90, 89, 87, 3, 7, 87, 92, 99, 4, 7
What I would like to do is summarize the info in the following manner:
ID, Name, Sex, Youth, Count(WeekNo), Sum(A1+A2+A3+B1+B2+B3), Sum(A2X+A3X+B2X+B3X) such that I would get the following results:
10, John Smith, M, , 4, 2176, 84
23, Jane Doe, F, , 5, 2720, 105
67, Johnny Jones, M, Y, 3, 1632, 63
I have been able to get the summed part done, but my attempts to do the count have failed. I know that the resulting query will be straight forward, but right now I'm cross-eyed.
Thanks.
-BT.
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