self-made Split

[Kaverin]

Recently someone asked me about using Split, and neither I nor this person had VB6, so I decided to make up something of my own that does the same thing. I'll show my function below for any who might be interested (really only applies to VB5 or lower users). Also, is this similar to how Split really operates in VB6? If anyone has any ideas for improvements, or wants to point out my sloppiness , please let me know. This is just a working version. One problem I know of already is that it won't handle the token delimiter appearing consecutively. Like if you have 2 or 3 spaces between words, and use a space as the token delimiter. If there are an even number of spaces, one will be tacked onto the next token found. For odd numbers of spaces, a space appears as a token itself.

Public Function Split(strString As String, strTokenDelimiter As String, strTokens() As String) As Long
   
   Dim lngCurPos As Long
   Dim lngLastPos As Long
   Dim lngNumTokens As Long
   
   If (strString = "") Or (strTokenDelimiter = "") Then
      Split = 0
      
      Exit Function
   End If
   
   If InStr(1, strString, strTokenDelimiter, vbTextCompare) = 0 Then
      lngNumTokens = 1
      
      ReDim strTokens(lngNumTokens) As String
      
      strTokens(lngNumTokens - 1) = strString
      
      Split = lngNumTokens
      
      Exit Function
   End If
   
   lngLastPos = 1
   lngCurPos = InStr(lngLastPos, strString, strTokenDelimiter, vbTextCompare)
   lngNumTokens = 0
   
   Do
      lngNumTokens = lngNumTokens + 1
      
      ReDim Preserve strTokens(lngNumTokens) As String
      
      strTokens(lngNumTokens - 1) = Mid$(strString, lngLastPos, lngCurPos - lngLastPos)
      
      lngLastPos = lngCurPos + 1
      lngCurPos = InStr(lngLastPos + 1, strString, strTokenDelimiter, vbTextCompare)
   
   Loop Until lngCurPos = 0
   
   lngNumTokens = lngNumTokens + 1
   
   ReDim Preserve strTokens(lngNumTokens) As String
   
   strTokens(lngNumTokens - 1) = Mid$(strString, lngLastPos)
   
   Split = lngNumTokens

End Function

Basically, you pass along a string to parse, the token delimiter of the string, and an array to hold the tokens (if found). The return value is the number of total tokens found, or 0 if there were none. I set it up so it always finds 1 token (that token being the original string itself) as long as both the string to parse and the token delimiter are valid strings themselves. Otherwise, it will split up the string and stick the tokens into the array. The array will go from 0 to lngNumTokens - 1 (I start my arrays at 0 as a result of old C habits )

[Edited by Kaverin on 11-02-2000 at 03:19 PM]