String Manipulation!!

[MrVB50au]

I have a string in which I want to cut off the last 3 characters and add 3 new characters.

EXAMPLE:
From:
C:\Windows\Temp\Demo.Txt
To:
C:\Windows\Temp\Demo.New

I know this is a simple task but for the life of me I have completely forgotten.

If anyone could help me out, this would be very much appreciated.

Thanks,

Andrew.

 

[DrJavaJoe]

Try this:

strNew = Left(strOld, Len(strOld) - 3) & "new"

"Two strings walk into a bar. The first string says to the bartender: 'Bartender, I'll have a beer. u.5n$x5t?*&4ru!2[sACC~ErJ'. The second string says: 'Pardon my friend, he isn't NULL terminated'."

 

[vb5prgrmr]

DrJavaJoe,...

While that may work for files with just 3 letters for an extension what about files that have 4 letter extensions (jpeg)?
[tt]
S = Left(S, InStrRev(S, "."))
[/tt]

Just a more versatile version.

Good Luck

 

[DrJavaJoe]

Good point I guess you would also want to make sure that it had an extension in the first place.

If InStr(s, ".") <> 0 Then
s = Left(s, InStrRev(s, ".")) & "new"
Else
s = s & ".new"
End If

&quot;Two strings walk into a bar. The first string says to the bartender: 'Bartender, I'll have a beer. u.5n$x5t?*&4ru!2[sACC~ErJ'. The second string says: 'Pardon my friend, he isn't NULL terminated'.&quot;

 

[vb5prgrmr]

True, very true. Good catch

 

[DrJavaJoe]

Where's strongm with his RegExp solution?

&quot;Two strings walk into a bar. The first string says to the bartender: 'Bartender, I'll have a beer. u.5n$x5t?*&4ru!2[sACC~ErJ'. The second string says: 'Pardon my friend, he isn't NULL terminated'.&quot;

 

[strongm]

Oh...hush up...

Private Function ChangeExtension(strSource As String, strNewExt As String) As String
    With CreateObject("VBScript.RegExp")
        .Pattern = "(^.+\.)(\w+)($)"
        ChangeExtension = .Replace(strSource, "$1" & strNewExt & "$3")
    End With
End Function