PowerShell

Get ZIP Directory Information as XML

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This example demonstrates how to use the GetDirectoryAsXML method to retrieve information about the contents of a ZIP archive as an XML document.

The returned XML describes the files currently contained in the ZIP object

PowerShell

Add-Type -Path "C:\chilkat\ChilkatDotNet47-x64\ChilkatDotNet47.dll"

$success = $false

$success = $false

# Open an existing ZIP archive.
$zip = New-Object Chilkat.Zip

$success = $zip.OpenZip("example.zip")
if ($success -eq $false) {
    $($zip.LastErrorText)
    exit
}

# Get the ZIP directory information as XML.
$xml = $zip.GetDirectoryAsXML()

$($xml)

# Suppose the ZIP contains:

# data/config/settings.json
# docs/readme.txt
# images/logo.png

# The XML contains one element for each ZIP entry.
# Example:
# 
# <?xml version="1.0" encoding="utf-8"?>
# <zip_contents>
#     <dir name="data">
#         <dir name="config">
#             <file>settings.json</file>
#         </dir>
#     </dir>
#     <dir name="docs">
#         <file>readme.txt</file>
#     </dir>
#     <dir name="images">
#         <file>logo.png</file>
#     </dir>
# </zip_contents>

$zip.CloseZip()

$("Done.")