Lua Requires Chilkat v11.0.0+

Get ZIP Directory Information as XML

See more Zip Examples

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

Lua

local chilkat = require("chilkat")

-- Open an existing ZIP archive.
local zip = chilkat.Zip.new()

local success = zip:open_zip("example.zip")
if not success then
    print(zip.last_error_text)
    return
end

-- Get the ZIP directory information as XML.
local xml = zip:get_directory_as_xml()

print(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:close_zip()

print("Done.")