Count occurrences of a character in a string

Question

Looking for the best way to do this in VB6. Typically, I would use this approach...

   ' count spaces
    For i = 1 To Len(text)
        If Mid$(text, i, 1) = " " Then count = count + 1 
    Next

Answer 1

Not saying it's the best way, but you code do:

distinctChr = " "
count = Len(text) - Len(Replace(text, distinctChr , ""))

Answer 2

Use the split command like this

Dim TempS As String
TempS = " This is a split  test "
Dim V As Variant
V = Split(TempS, " ")
Cls
Print UBound(V) '7
V = Split(TempS, "i")
Print UBound(V) '3
V = Split(TempS, "e")
Print UBound(V) '1

You can combine it to a single line.

Print UBound(Split(TempS, "i"))

I did some crude timing on it. On a 40,000 character string with all spaces it seems to clock in at 17 milliseconds on a 2.4 GHz Intel Core 2 processor.

A function could look like this

Function CountChar(ByVal Text As String, ByVal Char As String) As Long
    Dim V As Variant
    V = Split(Text, Char)
    CountChar = UBound(V)
End Function