(Unicode C++) 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

Note: This example requires Chilkat v11.0.0 or greater.

Chilkat C/C++ Library Downloads
MS Visual C/C++ C++ Builder Linux C/C++ Alpine Linux C/C++	MacOS C/C++ iOS C/C++ Android C/C++ MinGW C/C++

#include <CkZipW.h>

void ChilkatSample(void)
    {
    bool success = false;

    success = false;

    // Open an existing ZIP archive.
    CkZipW zip;

    success = zip.OpenZip(L"example.zip");
    if (success == false) {
        wprintf(L"%s\n",zip.lastErrorText());
        return;
    }

    // Get the ZIP directory information as XML.
    const wchar_t *xml = zip.getDirectoryAsXML();

    wprintf(L"%s\n",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();

    wprintf(L"Done.\n");
    }