Getting filename out of a full path

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March 22nd, 2006, 03:23 AM

ssimons

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Getting filename out of a full path

Hi Guys,

I'm using a vbs script for postprocessing image files.
The application is use for opening the images call's a vbs file with "filename" as a parameter.

Now the parameter "filename" contains the FULL path to the file. Since i need only the filename for some puposes is use a function to strip the filename out of the full path:

Code:

Function GetFileName(Path)
		Path = Replace(Path, "/", "\")
		If (Len(Path) > 0) AND (InStrRev(Path, "\") <> Len(Path)) Then
			GetFileName = Right(Path, (InStrRev(Path, "\")))
		Else
			Exit Function
		End If

if i have this path: d:\test files\SS_9__IMG12535.jpg
the function will translate this into: 2535.jpg

Can anyone help me with this?

Thx!

March 22nd, 2006, 03:26 AM

nofriends

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Hi,
try using the split function

Code:

Function GetFileName(Path)
		Path = Replace(Path, "/", "\")
		
		tempPath = split(Path, "\")
		If Len(tempPath(uBound(tempPath))) > 0  Then
			GetFileName = tempPath(uBound(tempPath))
		Else
			Exit Function
		End If

hope this helps

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March 22nd, 2006, 03:28 AM

ssimons

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Join Date: Jan 2006

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THX M8!

This seems work fine for me!

March 22nd, 2006, 03:33 AM

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no problem, glad its working

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