Java

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

Java

import com.chilkatsoft.*;

public class ChilkatExample {

  static {
    try {
        System.loadLibrary("chilkat");
    } catch (UnsatisfiedLinkError e) {
      System.err.println("Native code library failed to load.\n" + e);
      System.exit(1);
    }
  }

  public static void main(String argv[])
  {
    boolean success = false;

    //  Open an existing ZIP archive.
    CkZip zip = new CkZip();

    success = zip.OpenZip("example.zip");
    if (success == false) {
        System.out.println(zip.lastErrorText());
        return;
        }

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

    System.out.println(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();

    System.out.println("Done.");
  }
}