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Tip of the Day
Language: VB4,VB5
Expertise: Intermediate
May 15, 1999

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Replace - String replacement under VB4 and VB5

' A clone of the VB6's Replace function for use under VB5
' Note: many routines in this collection uses the Replace function

Function Replace(Source As String, Find As String, ReplaceStr As String, _
    Optional ByVal Start As Long = 1, Optional Count As Long = -1, _
    Optional Compare As VbCompareMethod = vbBinaryCompare) As String

    Dim findLen As Long
    Dim replaceLen As Long
    Dim index As Long
    Dim counter As Long
    
    findLen = Len(Find)
    replaceLen = Len(ReplaceStr)
    ' this prevents an endless loop
    If findLen = 0 Then Err.Raise 5
    
    If Start < 1 Then Start = 1
    index = Start
    
    ' let's start by assigning the source to the result
    Replace = Source
    
    ' if Find and ReplaceStr strings have same length, it is possible to
    ' use an optimized algorithm, based on the Mid$ command
    Do
        index = InStr(index, Replace, Find, Compare)
        If index = 0 Then Exit Do
        If findLen = replaceLen Then
            ' if the find and replace strings have same length
            ' we can use the faster Mid$ command
            Mid$(Replace, index, findLen) = ReplaceStr
        Else
            ' else we must use concatenation
            Replace = Left$(Replace, index - 1) & ReplaceStr & Mid$(Replace, _
                index + findLen)
        End If
        ' skip over the string just added
        index = index + replaceLen
        ' increment the replacement counter
        counter = counter + 1
        ' Note that the Loop Until test will always fail if Count = -1
    Loop Until counter = Count
    
    ' The next operation serves to keep complete compatibility with
    ' VB6's Replace function. You can delete it if you prefer.
    If Start > 1 Then Replace = Mid$(Replace, Start)

End Function
Francesco Balena
 
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