Get Filename And Path From A Pathfilename Spec

[shayneh]

I would like to get the path and filename from a pathfilename spec that doesn't necessarily exist on the file system.

In vb script:

Set fso = CreateObject("Scripting.FileSystemObject")
   file = fso.GetFileName(pathfilename)

How can this be done in AutoIt3?

Thanks,

ShayneH

Edited April 14, 2006 by shayneh

[cdkid]

Umm, I'm not sure what exactly the Scripting.FileSystemObject.GetFileName does. I'm guessing it gives you the filename suchas "text.txt" and the path might be "c:\program files". Try something like

$file = "c:\Path\To\The\File.txt"
$file = StringSplit($file, "\")
$path = ""
For $i = 1 To $file[0] - 1
$path = $path & $file[$i] & "\"
next
$filename = $file[$file[0]]
msgbox(0,'',$path & $filename)

Or you can use the same object approach you've been using

$fso = ObjCreate("Scripting.FileSystemObject")
$a = $fso.GetFileName($SomeVariableHoldingTheFile)

~cdkid

[Jos]

-or-

Standard UDF: _PathSplit()

[shayneh]

Thanks

[Simucal]

$filepath = "c:\Path\To\The\File.txt"

$file = StringTrimLeft($filepath,StringInStr($filepath,"\",0,-1))

Pure Genius Larry! I've always used:

$ShortFilePathArray = StringSplit($FilePath,"\")
$ShortFilePath = _ArrayPop($ShortFilePathArray)

[hoppy]

$filepath = "c:\Path\To\The\File.txt"

$file = StringTrimLeft($filepath,StringInStr($filepath,"\",0,-1))

Nice and short and efficiant.

How can I get only the path like this?

[Jos]

Nice and short and efficiant.

How can I get only the path like this?

$filepath = "c:\Path\To\The\File.txt"
$file = StringTrimLeft($filepath,StringInStr($filepath,"\",0,-1))
$Path = StringLeft($filepath,StringInStr($filepath,"\",0,-1)-1)

[hoppy]

$filepath = "c:\Path\To\The\File.txt"
$file = StringTrimLeft($filepath,StringInStr($filepath,"\",0,-1))
$Path = StringLeft($filepath,StringInStr($filepath,"\",0,-1)-1)

Thanks a lot

Was making thing to complicated by lack of knowledge of all the possible commands.