How to unzip a file in Powershell?

Question

I have a .zip file and need to unpack its entire content using Powershell. I'm doing this but it doesn't seem to work:

$shell = New-Object -ComObject shell.application
$zip = $shell.NameSpace("C:\a.zip")
MkDir("C:\a")
foreach ($item in $zip.items()) {
  $shell.Namespace("C:\a").CopyHere($item)
}

What's wrong? The directory C:\a is still empty.

Answer 1

In PowerShell v5+, there is an Expand-Archive command (as well as Compress-Archive) built in:

Expand-Archive C:\a.zip -DestinationPath C:\a

Answer 2

Here is a simple way using ExtractToDirectory from System.IO.Compression.ZipFile:

Add-Type -AssemblyName System.IO.Compression.FileSystem
function Unzip
{
    param([string]$zipfile, [string]$outpath)

    [System.IO.Compression.ZipFile]::ExtractToDirectory($zipfile, $outpath)
}

Unzip "C:\a.zip" "C:\a"

Note that if the target folder doesn't exist, ExtractToDirectory will create it. Other caveats:

• Existing files will not be overwritten and instead trigger an IOException.
• This method requires at least .NET Framework 4.5, available for Windows Vista and newer.
• Relative paths are not resolved based on the current working directory, see Why don't .NET objects in PowerShell use the current directory?

See also:

• How to Compress and Extract files (Microsoft Docs)

Answer 3

In PowerShell v5.1 this is slightly different compared to v5. According to MS documentation, it has to have a -Path parameter to specify the archive file path.

Expand-Archive -Path Draft.Zip -DestinationPath C:\Reference

Or else, this can be an actual path:

Expand-Archive -Path c:\Download\Draft.Zip -DestinationPath C:\Reference

Expand-Archive Doc

Answer 4

Use Expand-Archive cmdlet with one of parameter set:

Expand-Archive -LiteralPath C:\source\file.Zip -DestinationPath C:\destination

Expand-Archive -Path file.Zip -DestinationPath C:\destination

Answer 5

Hey Its working for me..

$shell = New-Object -ComObject shell.application
$zip = $shell.NameSpace("put ur zip file path here")
foreach ($item in $zip.items()) {
  $shell.Namespace("destination where files need to unzip").CopyHere($item)
}

Answer 6

Use the built in powershell method Expand-Archive

Example

Expand-Archive -LiteralPath C:\archive.zip -DestinationPath C:\

Answer 7

Using expand-archive but auto-creating directories named after the archive:

function unzip ($file) {
    $dirname = (Get-Item $file).Basename
    New-Item -Force -ItemType directory -Path $dirname
    expand-archive $file -OutputPath $dirname -ShowProgress
}

Answer 8

For those, who want to use Shell.Application.Namespace.Folder.CopyHere() and want to hide progress bars while copying, or use more options, the documentation is here:
https://learn.microsoft.com/en-us/windows/desktop/shell/folder-copyhere

To use powershell and hide progress bars and disable confirmations you can use code like this:

# We should create folder before using it for shell operations as it is required
New-Item -ItemType directory -Path "C:\destinationDir" -Force

$shell = New-Object -ComObject Shell.Application
$zip = $shell.Namespace("C:\archive.zip")
$items = $zip.items()
$shell.Namespace("C:\destinationDir").CopyHere($items, 1556)

Limitations of use of Shell.Application on windows core versions:
https://learn.microsoft.com/en-us/windows-server/administration/server-core/what-is-server-core

On windows core versions, by default the Microsoft-Windows-Server-Shell-Package is not installed, so shell.applicaton will not work.

note: Extracting archives this way will take a long time and can slow down windows gui

Answer 9

function unzip {
    param (
        [string]$archiveFilePath,
        [string]$destinationPath
    )

    if ($archiveFilePath -notlike '?:\*') {
        $archiveFilePath = [System.IO.Path]::Combine($PWD, $archiveFilePath)
    }

    if ($destinationPath -notlike '?:\*') {
        $destinationPath = [System.IO.Path]::Combine($PWD, $destinationPath)
    }

    Add-Type -AssemblyName System.IO.Compression
    Add-Type -AssemblyName System.IO.Compression.FileSystem

    $archiveFile = [System.IO.File]::Open($archiveFilePath, [System.IO.FileMode]::Open)
    $archive = [System.IO.Compression.ZipArchive]::new($archiveFile)

    if (Test-Path $destinationPath) {
        foreach ($item in $archive.Entries) {
            $destinationItemPath = [System.IO.Path]::Combine($destinationPath, $item.FullName)

            if ($destinationItemPath -like '*/') {
                New-Item $destinationItemPath -Force -ItemType Directory > $null
            } else {
                New-Item $destinationItemPath -Force -ItemType File > $null

                [System.IO.Compression.ZipFileExtensions]::ExtractToFile($item, $destinationItemPath, $true)
            }
        }
    } else {
        [System.IO.Compression.ZipFileExtensions]::ExtractToDirectory($archive, $destinationPath)
    }
}

Using:

unzip 'Applications\Site.zip' 'C:\inetpub\wwwroot\Site'

Answer 10

ForEach Loop processes each ZIP file located within the $filepath variable

    foreach($file in $filepath)
    {
        $zip = $shell.NameSpace($file.FullName)
        foreach($item in $zip.items())
        {
            $shell.Namespace($file.DirectoryName).copyhere($item)
        }
        Remove-Item $file.FullName
    }