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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

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Local $bSuccess = False

$bSuccess = False

; Open an existing ZIP archive.
$oZip = ObjCreate("Chilkat.Zip")

$bSuccess = $oZip.OpenZip("example.zip")
If ($bSuccess = False) Then
    ConsoleWrite($oZip.LastErrorText & @CRLF)
    Exit
EndIf

; Get the ZIP directory information as XML.
Local $sXml = $oZip.GetDirectoryAsXML()

ConsoleWrite($sXml & @CRLF)

; 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>

$oZip.CloseZip 

ConsoleWrite("Done." & @CRLF)