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Home/ Questions/Q 5984463
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Editorial Team
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Editorial Team
Asked: May 22, 20262026-05-22T22:22:45+00:00 2026-05-22T22:22:45+00:00

I’m trying to zip up log using DotNetZip and powershell. The files are in

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I’m trying to zip up log using DotNetZip and powershell. The files are in C:\user\temp\logs When I loop through the logs in the directory and add them to the zip file, I end up with the folder hierarchy and the log files when I only want the log files.

So the zip ends up containing:

-user
  └temp  
    └logs
       └log1.log
        log2.log
        log3.log

When I want the zip file to contain is:

log1.log
log2.log
log3.log

Here is the script I’m running to test with:

[System.Reflection.Assembly]::LoadFrom("c:\\\User\\bin\\Ionic.Zip.dll");
$zipfile = new-object Ionic.Zip.ZipFile("C:\user\temp\logs\TestZIP.zip");

$directory = "C:\user\temp\logs\"
$children = get-childitem -path $directory
foreach ($o in $children)
{
   if($o.Name.EndsWith(".log")){
      $e = $zipfile.AddFile($o.FullName)
   }
}
$zipfile.Save()
$zipfile.Dispose()
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-22T22:22:46+00:00Added an answer on May 22, 2026 at 10:22 pm

    There is an AddFile where you can override the filename in the archive:

    public ZipEntry AddFile(
        string fileName,
        string directoryPathInArchive
    )
    

    fileName (String)

    The name of the file to add. The name of the file may be a relative
    path or a fully-qualified path.

    directoryPathInArchive (String)

    Specifies a directory path to use to override any path in the fileName.
    This path may, or may not, correspond
    to a real directory in the current
    filesystem. If the files within the
    zip are later extracted, this is the
    path used for the extracted file.
    Passing null (Nothing in VB) will use
    the path on the fileName, if any.
    Passing the empty string (“”) will
    insert the item at the root path
    within the archive.

    Try this:

     $e = $zipfile.AddFile($o.FullName, $o.Name)
    

    It is also possible that this does what you want:

     $e = $zipfile.AddFile($o.FullName, "")
    
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