by Chris Huff (2 Submissions)
Category: String Manipulation
Compatability: VB 6.0
Difficulty: Unknown Difficulty
Originally Published: Tue 20th April 2004
Date Added: Mon 8th February 2021
Rating: 
(1 Votes)
Counts the number of occurrences of a character/character string. I found one code example on the net and it took 4 milliseconds to run in a
VBcoders
'txtCount = shows # occurances
'txtSearch = multi-line textbox that holds the contents of what you need to scan
'txtWord = the character/character string to count.
'Look at your immediate window to see the time of running, in milliseconds.
Private Sub Command1_Click()
Debug.Print MyTime
txtCount = CharCount2(txtSearch, txtWord)
Debug.Print MyTime
End sub
Public Function MyTime() As String
MyTime = Format(Now, "dd-MMM-yyyy HH:nn:ss") & "." & Right(Format(Timer, "#0.00"), 2)
End Function
Public Function CharCount2(OrigString As String, Chars As String) As Long
'get number of occurences
CharCount2 = 0
On Error GoTo errFunc
Dim lCharLen As Long
Dim lAns As Long
Dim sInput As String
Dim sChar As String
Dim isFound As Boolean
Dim lngStart As Long
If OrigString = "" Then Exit Function
If Chars = "" Then Exit Function
sInput = OrigString
sInput = LCase(sInput)
sChars = LCase(Chars)
lCharLen = Len(sChars)
isFound = True
lngStart = 1
Do Until isFound = False
x = InStr(lngStart, sInput, sChars)
If x > 0 Then
lAns = lAns + 1
lngStart = x + lCharLen
Else
isFound = False
End If
Loop
CharCount2 = lAns
Exit Function
errFunc:
Exit Function
End Function
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