MS Access VBA InStrRev question

InStrRev is correct. In the first case, "\" is the 8th character from the front. In the second case, it in the 9th character from the front. InStrRev searches from back to front but returns the character position count from the front. If you went the count from the back use:
Len (YourString) - InStrRev("YourString", "\")

Same here.
Adding lots of . into the string doesn't change the result. Looks like a bug at the moment.

here's the function I'm using this in.  

Input
?returnfilepath("C:\test\te.st.txt")
 17
 8
C:\test\t

?returnfilepath("C:\te.st\te.st.txt")
 18
 9
C:\te.st\

Why am I getting the extra "t" in the first one

Public Function ReturnFilePath(Filename As Variant)
    'returns the path
    Dim ln As Integer
    If Filename <> "" Then
        ln = Len(Filename)
        Debug.Print ln
        Debug.Print InStrRev(Filename, "\")
        ReturnFilePath = Left(Filename, ln - InStrRev(Filename, "\"))
    End If
End Function

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And it looks like I can't count- back to sleep Peter!

Try this function


Public Function getpath(FileName As String) As String
getpath = (Mid(FileName, 1, Len(FileName) - Len(Dir(FileName))))
End Function

Hello keschuster,

You are getting confused.  The len - position is a requirement of the right method.  For lewft you want the position of the "\" to the left therefore use:

Regards,

chris_bottomley

Public Function ReturnFilePath(Filename As Variant)
    'returns the path
    Dim ln As Integer
    If Filename <> "" Then
        ln = Len(Filename)
        Debug.Print ln
        Debug.Print InStrRev(Filename, "\")
        ReturnFilePath = Left(Filename, InStrRev(Filename, "\") - 1)
    End If
End Function

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TheNelson
Not working it's returning

C:\test\te.st.txt
C:\te.st\te.st.txt

Need everything after \ stripped off

Chris
Test both inputs

C:\test\te.st.txt
C:\te.st\te.st.txt

two different results