Get parent folder

[DTNGUYEN]

I would like to get parent folder path base on @ScriptDir. Please kind to help how to do it.

Eg, @ScriptDir returns "C:\AutoIt\Script

I would like to get "C:\AutoIt" in this case.

Edited April 12, 2010 by DTNGUYEN

[ChrisL]

$ParentDir = StringLeft(@scriptDir,StringInStr(@scriptDir,"\",0,-1)-1)

Msgbox(0,"", $ParentDir )

[KaFu]

$dir = @ScriptDir
$dir_parent = StringLeft($dir,StringInStr($dir,"\",0,-1)-1)
MsgBox(0,"",$dir_parent)

[Yashied]

StringRegExpReplace(@ScriptDir, '\\[^\\]*$', '')

Edited April 12, 2010 by Yashied

[DTNGUYEN]

Thanks for help. However, i just find out another way to get by using _PathFull

#incllude<file.au3>

@LIB_DIR = _PathFull(@ScriptDir & "..\..\") & "LIB\"

[Cyri]

StringRegExpReplace(@ScriptDir, '\\[^\\]*$', '')

Slightly off topic...I like this solution, but how would I modify it to only show the parent folder name and trim the left side of the path off. For example:

Have this:

$sPath = "C:\test1\test2\test3\test4\test.xml"

Need this:

$sParent = "test4"

Basically just the parent folder name and not the full path to the parent folder.

[sahsanu]

Have this:

$sPath = "C:\test1\test2\test3\test4\test.xml"

Need this:

$sParent = "test4"

Basically just the parent folder name and not the full path to the parent folder.

Something like this?

$test=StringRegExpReplace(@ScriptDir, '\\[^\\]*$', '')
$test=StringRegExpReplace($test, '.*\\', '')

[Cyri]

Something like this?

$test=StringRegExpReplace(@ScriptDir, '\\[^\\]*$', '')
$test=StringRegExpReplace($test, '.*\\', '')

Thanks that should do the trick.