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Help with MS Access query

Posted on 2013-06-20
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Last Modified: 2013-11-19
I need a query that delineates new records that have been added for each month.  The records have a unique ID.  The update month is also a field. Please see the example data database and the desired results (excel)

Any Takers!
TestMDB.accdb
Example.xlsx
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Question by:Thomask23
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4 Comments
 
LVL 24

Expert Comment

by:Bitsqueezer
ID: 39264262
Hi,

maybe this query is what you want:

TRANSFORM Count([Sample Table].[ID]) AS AnzahlvonID
SELECT 'Count Of Dates' AS [Count of Dates], Count([Sample Table].[ID]) AS [Count of all]
FROM [Sample Table]
GROUP BY 'Count Of Dates'
PIVOT Format([Update Date],"Short Date");

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You can of course also group that by "Data":

TRANSFORM Count([Sample Table].ID) AS AnzahlvonID
SELECT [Sample Table].Data, Count([Sample Table].ID) AS [Count of all]
FROM [Sample Table]
GROUP BY [Sample Table].Data
PIVOT Format([Update Date],"Short Date");

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Cheers,

Christian
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Author Comment

by:Thomask23
ID: 39265740
Almost...

The records are identified as new by the ID. If a new id is updated to the data it should only be count it. I want to count the only the new records added by the updated date.
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Accepted Solution

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Bitsqueezer earned 500 total points
ID: 39267402
Hi,

that was not clear from your example and to be exact, it is not clear from what you wrote now.

You can create a query like this to get the newest ID of each date if that is what you want:

SELECT [Sample Table].[Update Date], Max([Sample Table].ID) AS MaxvonID
FROM [Sample Table]
GROUP BY [Sample Table].[Update Date];

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You see that any date value always only have one ID and so the result of a count in a further query would always be 1.

(That can also be transformed like in the example above but normally it's better to create normal queries than crosstab queries as these have a limitation of 255 columns.)

If that's not what you thought of then please be more exact.

A "new ID" cannot be "updated", only inserted.

Cheers,

Christian
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Author Comment

by:Thomask23
ID: 39659432
Thanks
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