是否可以使用PowerShell创建zip存档?


当前回答

这里有一个完整的命令行示例,可以从cmd.exe或ssh或您想要的任何地方启动!

powershell.exe -nologo -noprofile -command "&{ Add-Type -A 'System.IO.Compression.FileSystem' [System.IO.Compression.ZipFile]::CreateFromDirectory('c:/path/to/source/folder/', 'c:/path/to/output/file.zip');}"

问候

其他回答

一个纯PowerShell的替代方案,与PowerShell 3和。net 4.5(如果你可以使用它):

function ZipFiles( $zipfilename, $sourcedir )
{
   Add-Type -Assembly System.IO.Compression.FileSystem
   $compressionLevel = [System.IO.Compression.CompressionLevel]::Optimal
   [System.IO.Compression.ZipFile]::CreateFromDirectory($sourcedir,
        $zipfilename, $compressionLevel, $false)
}

只需传入您想要创建的zip归档文件的完整路径,以及包含您想要压缩的文件的目录的完整路径。

System.IO.Packaging.ZipPackage呢?

它需要。net 3.0或更高版本。

#Load some assemblys. (No line break!)
[System.Reflection.Assembly]::Load("WindowsBase, Version=3.0.0.0, Culture=neutral, PublicKeyToken=31bf3856ad364e35")

#Create a zip file named "MyZipFile.zip". (No line break!)
$ZipPackage=[System.IO.Packaging.ZipPackage]::Open("C:\MyZipFile.zip",
   [System.IO.FileMode]"OpenOrCreate", [System.IO.FileAccess]"ReadWrite")

#The files I want to add to my archive:
$files = @("/Penguins.jpg", "/Lighthouse.jpg")

#For each file you want to add, we must extract the bytes
#and add them to a part of the zip file.
ForEach ($file In $files)
{
   $partName=New-Object System.Uri($file, [System.UriKind]"Relative")
   #Create each part. (No line break!)
   $part=$ZipPackage.CreatePart($partName, "",
      [System.IO.Packaging.CompressionOption]"Maximum")
   $bytes=[System.IO.File]::ReadAllBytes($file)
   $stream=$part.GetStream()
   $stream.Write($bytes, 0, $bytes.Length)
   $stream.Close()
}

#Close the package when we're done.
$ZipPackage.Close()

通过安德斯·赫塞尔博姆

如果你安装了WinRAR:

function ZipUsingRar([String] $directory, [String] $zipFileName)
{
  Write-Output "Performing operation ""Zip File"" on Target ""Item: $directory Destination:"
  Write-Output ($zipFileName + """")
  $pathToWinRar = "c:\Program Files\WinRAR\WinRar.exe";
  [Array]$arguments = "a", "-afzip", "-df", "-ep1", "$zipFileName", "$directory";
  & $pathToWinRar $arguments;
}

参数的含义:afzip创建zip归档,df删除文件,ep1不创建归档内的完整目录路径

旧线程,但仍然,我到了这里:)

这不是最初问题的答案,但也许有人会发现如何用PS创建ZipArchive对象是有用的。

# Example, if you have like I have a $myByteArray byte[] with the compressed data:
Add-Type -AssemblyName System.IO.Compression.FileSystem

# Fixed length stream:
$strm = New-Object -TypeName System.IO.MemoryStream -ArgumentList @(, $myByteArray);

# Create ZipArchive object
$arch = [System.IO.Compression.ZipArchive]::new($strm);

# List entries
$arch.Entries

下面是工作代码,压缩源文件夹中的所有文件,并在目标文件夹中创建一个zip文件。

    $DestZip="C:\Destination\"
    $Source = "C:\Source\"

    $folder = Get-Item -Path $Source

    $ZipTimestamp = Get-Date -format yyyyMMdd-HHmmss;
    $ZipFileName  = $DestZip + "Backup_" + $folder.name + "_" + $ZipTimestamp + ".zip" 

    $Source

    set-content $ZipFileName ("PK" + [char]5 + [char]6 + ("$([char]0)" * 18)) 
    # Wait for the zip file to be created.
    while (!(Test-Path -PathType leaf -Path $ZipFileName))
    {    
        Start-Sleep -Milliseconds 20
    } 
    $ZipFile = (new-object -com shell.application).NameSpace($ZipFileName)

    Write-Output (">> Waiting Compression : " + $ZipFileName)       

    #BACKUP - COPY
    $ZipFile.CopyHere($Source) 

    $ZipFileName
    # ARCHIVE

    Read-Host "Please Enter.."