Fields_A, Fields_B, Fields_C, Fied_D
        3           4               6            6

need to count each number
how can i do that ??
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Rey Obrero (Capricorn1)Commented:
what result do expect to see?
Rey Obrero (Capricorn1)Commented:
what result do you expect to see?
Dale FyeOwner, Developing Solutions LLCCommented:
What happens when one of the fields is NULL?
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Dale FyeOwner, Developing Solutions LLCCommented:
I would create a function which accepts a parameter array, pass the field values to the function and let the function loop through the elements of the parameter array.  If the value is numeric, increment a counter, if not, skip to the next element of the array.

You would call this something like:

NumCount: fnCountNumbers([Field_A], [Field_B], [Field_C], [Field_D])

Because this function would use a parameter array, you could pass as many columns worth of data as you want.
PortletPaulEE Topic AdvisorCommented:
is this what you are after?

Fields_A, Fields_B, Fields_C, Field_D, Result?
        3           4               6            6          4
        3           4               6         NULL      3
        3           4             NULL    NULL     2

if not please supply the "expected result" from a larger sample of data
MIKE NIKEAuthor Commented:
thnks all !

my data fiels is shorttxt format

i need count numbers in this way

Fields_A, Fields_B, Fields_C, Fied_D         1    2    3   4   5   6  7   8   9   0
        3           4               6            6                            1   1        2
        2          4                0           0                        1          1                          2
       1          2                 6           1                 2     1                  1

need  it to be   count how many times the number repeat
for each number
Rey Obrero (Capricorn1)Commented:
see this sample db.
open query Q_countnumbers

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MIKE NIKEAuthor Commented:

but i will need help please to put to work in
the database

how i do that ?
Rey Obrero (Capricorn1)Commented:
- from your database, import the module modCountNumbers from the sample db
- create a query similar to the query Q_count using your table.
Here's a version that uses the Regexp object to count the instances:
Function MatchCount(ByVal parmConcatFields As String, parmMatch, Optional parmDelim As String = ",") As Long
    Static oRE As Object
    If oRE Is Nothing Then
        Set oRE = CreateObject("vbscript.regexp")
        oRE.Global = True
    End If
    oRE.pattern = "\b" & parmMatch & "(?:" & parmDelim & "|$)"
    If oRE.test(parmConcatFields) Then
        MatchCount = oRE.Execute(parmConcatFields).Count
        MatchCount = 0
    End If
End Function

Open in new window

You can simplify your SQL by concatenating the fields into a new field (named ConcatFields in the following example:
SELECT tableN.A, tableN.B, tableN.C, tableN.D, 
tableN.A & "," &  tableN.B & "," &  tableN.C & "," & tableN.D As ConcatFields,
CountOf(ConcatFields,1) AS 1, CountOf(ConcatFields,2) AS 2, CountOf(ConcatFields,3) AS 3, CountOf(ConcatFields,4) AS 4, CountOf(ConcatFields,5) AS 5, CountOf(ConcatFields,6) AS 6, CountOf(ConcatFields,7) AS 7, CountOf(ConcatFields,8) AS 8, CountOf(ConcatFields,9) AS 9, CountOf(ConcatFields,0) AS 0
FROM tableN;

Open in new window

When using the Regexp function, the query would look like this:
SELECT tableN.A, tableN.B, tableN.C, tableN.D, 
tableN.A & "," &  tableN.B & "," &  tableN.C & "," & tableN.D As ConcatFields,
MatchCount(ConcatFields,1) AS 1, MatchCount(ConcatFields,2) AS 2, MatchCount(ConcatFields,3) AS 3, MatchCount(ConcatFields,4) AS 4, MatchCount(ConcatFields,5) AS 5, MatchCount(ConcatFields,6) AS 6, MatchCount(ConcatFields,"7") AS 7, MatchCount(ConcatFields,"8") AS 8, MatchCount(ConcatFields,"9") AS 9, MatchCount(ConcatFields,"0") AS 0
FROM tableN;

Open in new window

Note: The MatchCount function can accept either numeric or string literals to match and can accept a different field delimiter character, as I illustrated with values 7-9,0 in the above example.
MIKE NIKEAuthor Commented:
thanks you lead the way !
MIKE NIKEAuthor Commented:
all of you thanks !!
Rey  Obrero keep the good job !
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