VBA Count number of different values within a column, starting from a fixed point, ending at last cell that has a value in it

user_19611116 0
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VBA Hey guys, I just started to work with VBA and i am looking for a solution to my problem. The task is to count the number of different Values (Suppliers) inside a column, starting at a fixed point, lets say for column A to start at A5, but end at the last value in that column. Every time i try to put the range from A5 to last written cell in column A I either get errors or a wrong answer. Anyone here that can help me out?

Sub Count_Values_()

Sheets("Präsentation").Select

Dim dblAnz As Double

Dim rngRange As Range, rngRangeCnt As Range

Set rngRange = Range("A5:A50")

For Each rngRangeCnt In rngRange
    dblAnz = dblAnz + 1 / WorksheetFunction.CountIf(rngRange, rngRangeCnt.Text)
Next


Dim Bereich  As Range

Dim Zelle    As Range

Dim lAnzahl  As Long

   Set Bereich = Range("A5:A50")
   
   For Each Zelle In Bereich
      If Zelle.Font.Bold = True Then
         lAnzahl = lAnzahl + 1
      End If
   Next Zelle

Sheets("Anleitung").Select

Range("F1").Value = dblAnz - lAnzahl - 1

End Sub

This is the code I used before by searching values from A5 to A50 and subtracting all the bold headlines. I also added a -1 as an empty cell counts as a value as well I guess or at least I got the right numbers this way.

vba excel
1 Answers

Please, try the next way:

Sub testCountDefinedRange()
   Dim sh As Worksheet, startRow As Long
   Const searchVal As Double = 25.13 'what is it to be searched for
   
   Set sh = ActiveSheet 'use here the sheet you need
   startRow = 5         'the starting row of the range to be searched

   Debug.Print WorksheetFunction.CountIf(sh.Range("A" & startRow, sh.Range("A" & sh.rows.count).End(xlUp)), searchVal)
End Sub

But according to community spirit, you should prove your involvement in solving the problem by your own. Even if you cannot show us a working piece of code/formula etc. Anything proving that you did something, at least, some research on the internet...

user_22333080
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