Solved How to unzip all the files in a particular path

July 8, 2014 at 01:10:51
Specs: Windows 7
Hi, I have used the below code for unzipping the files in VB but it is unzipping only when the file name is given in script but i need to zip all the files in a particular path ..Can you plz help on this?

'The location of the zip file.
ZipFile="C:\temp\test.zip"
'The folder the contents should be extracted to.
ExtractTo="C:\temp\new"

'Extract the contents of the zip file.
set objShell = CreateObject("Shell.Application")
set FilesInZip=objShell.NameSpace(ZipFile).items
objShell.NameSpace(ExtractTo).CopyHere(FilesInZip)
Set fso = Nothing
Set objShell = Nothing

message edited by Arunkumar11


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#1
July 13, 2014 at 11:04:59
✔ Best Answer
We need to get more VBScripters here, I swear.
Const zipPath = "C:\test"
Const extractTo = "C:\out\test"

Set fso = CreateObject("Scripting.FileSystemObject")
Set shell = CreateObject("Shell.Application")
Set zipFolder = shell.Namespace(zipPath)
Set destFolder = shell.Namespace(extractTo)
Set items = zipFolder.Items : items.Filter &H100E0, "*.zip"
For Each zip In Items
  destFolder.NewFolder fso.GetBaseName(zip.Path)
  shell.Namespace(fso.BuildPath(extractTo, fso.GetBaseName(zip.Path))) _
   .CopyHere shell.Namespace(zip).Items, &H214
Next 'zip
WScript.Echo "Done"

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message edited by Razor2.3


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