Zipping files with specific file name using powershell?

Question

I have a set of files named as: 20161205_abc, 20161205_bcd, 20161205_cde, 20161206_abc, 20161204_abc.

I have a script in place to zip all the files who have names like: 20161205*.

Add-Type -assembly 'System.IO.Compression'

[string]$zipFN = 'u:\users\riteshthakur\desktop\myZipFile.zip'
[string]$fileToZip = 'u:\users\riteshthakur\desktop\abc\20161205*.txt'

[System.IO.Compression.ZipArchive]$ZipFile = [System.IO.Compression.ZipFile]::Open($zipFN, ([System.IO.Compression.ZipArchiveMode]::Create))
[System.IO.Compression.ZipFileExtensions]::CreateEntryFromFile($ZipFile, $fileToZip, (Split-Path $fileToZip -Leaf))
$ZipFile.Dispose()

This throws error: "Exception calling "CreateEntryFromFile" with "3" argument(s): "Illegal characters in path." At U:\Users\riteshthakur\Desktop\Zip.PS1:7 char:1 + [System.IO.Compression.ZipFileExtensions]::CreateEntryFromFile($ZipFile, $fileTo ..."

Please help.

CreateEntryFromFile() takes exactly one file as its second argument, it won't treat * as a wildcard — Mathias R. Jessen
– Mathias R. Jessen, Commented Dec 12, 2016 at 13:57

BenH · Accepted Answer · 2016-12-12 15:10:28Z

1

Since * isn't an accepted character. You could replace your $filesToZip with a Get-ChildItem and a filter and then iterate through the values with a foreach.

Add-Type -assembly 'System.IO.Compression'
Add-Type -assembly 'System.IO.Compression.FileSystem'
[string]$zipFN = 'c:\temp\myZipFile.zip'
$filesToZip = (Get-ChildItem -Path "c:\temp\" -Filter test.*).fullname

[System.IO.Compression.ZipArchive]$ZipFile = [System.IO.Compression.ZipFile]::Open($zipFN,([System.IO.Compression.ZipArchiveMode]::Create))
foreach ($fileToZip in $filesToZip) {
    [System.IO.Compression.ZipFileExtensions]::CreateEntryFromFile($ZipFile, $fileToZip, (Split-Path $fileToZip -Leaf))
}
$ZipFile.Dispose()

edited Dec 12, 2016 at 15:10

answered Dec 12, 2016 at 14:29

BenH

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3 Comments

James Woolfenden Over a year ago

You're missing an import for [System.IO.Compression.ZipFile]

BenH Over a year ago

It was actually 'System.IO.Compression.FileSystem' that I was missing. And some stuff was in my loop that shouldnt have been.

Moose Over a year ago

Thanks a lot BenH. :)

Esperento57 · Accepted Answer · 2016-12-12 18:16:28Z

1

in powershell V5 you can simply do it like it:

$zipFN = 'u:\users\riteshthakur\desktop\myZipFile.zip'
$fileToZip = 'u:\users\riteshthakur\desktop\abc\20161205*.txt'
gci $fileToZip -File | Compress-Archive -DestinationPath $zipFN -Update

answered Dec 12, 2016 at 18:16

Esperento57

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1 Comment

Sergei Rodionov Over a year ago

Compress-Archive file limit is 2GB

js2010 · Accepted Answer · 2018-06-14 16:55:31Z

I like to preserve the folder structure. You have to be in the directory above what you want to zip for this script to work.

# myzip.ps1   
# example "ls -r foo | .\myzip foo.zip

# doesn't zip empty folders

Param([parameter(ValueFromPipeline=$True)]$file, 
  [parameter(position=0)]$zipfile)

Begin {
  if (test-path $zipfile) {
    'zipfile exists'
    exit 1
  }

  Add-Type -AssemblyName 'system.io.compression.filesystem'

  # update or create                                                            
  $zip = [System.IO.Compression.ZipFile]::Open($zipfile,'create')
}

Process {     
  # "test-path $file pathtype leaf" doesn't work right       
  if ($file | test-path -pathtype leaf) {
    $rel = $file | resolve-path -relative
    $rel = $rel -replace '^\.\\','' # take .\ off front                         
    $rel = $rel -replace '^\./',''  # osx take ./ off front
    # destination, sourcefilename, entryname, compressionlevel                  
    [System.IO.Compression.ZipFileExtensions]::CreateEntryFromFile($zip,
      $rel, $rel, 'optimal')
  } elseif ($file | test-path -pathtype container) {
    "$file is container"
  } else {
    "$file is unknown"
  }
}

End {
  $zip.dispose()
}

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