Count the number of times a value appears in a column using Ms Access

Hi experts,

How can we accomplish the following in Access.

Situation:
We have a column with an internal codes to identify trees. This internal code is unique for each tree.
But in cases where that tree is divided in 2 or 3 pieces, that internal code gets repeated in that column.

Now for reporting it is needed that we know how many trees there are and how many pieces.

How can we count each unique internal tree code?
I would like to see a column that shows for each record one of the following values: 1 or 0.5 or 0.33.

The values are determined by the counting of the internal tree code.
So if the counting result is
1, show value"1"
2, show value "0.5"
3, show value "0.33"

Am looking forward to a solution.
Shaun Kline

However, you may be able to use the replace function to "remove" the code from the string, then compare the length of the unaltered string against the length of the altered string to determine occurrences.

If your string where "alphabet", and you wanted to count the "a" letters in the string, you could do: Length("alphabet") - Length(Replace("alphabet", "a", "")).

For codes that are multiple letters, you would just need to divide the result of the above with the length of your code.
``````SELECT Count(Table1.tree) AS CountOftree, 10/([table1].[tree code]*10) AS Expr1
FROM Table1
GROUP BY 10/([table1].[tree code]*10);``````

Assuming Simple Table

Table1

Table1
tree codetree
1one
2two
3three
4four
2two again

Alright, the field "BoomCode" is the field with the unique internal tree code.

With your example, you can do a grouping on the BoomCode, and divide 1 by the count of occurrences.

Hey Shaun, yeah, I did the grouping of BoomCodes before I posted the question but was stuck there.

How do I do this " divide 1 by the count of occurrences."

Shaun Kline

membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.

Hey Shaun, awesome. Your suggestion works,

we are making progress. But not yet solved.