Delphi DLL

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

Delphi DLL

uses
    Winapi.Windows, Winapi.Messages, System.SysUtils, System.Variants, System.Classes, Vcl.Graphics,
    Vcl.Controls, Vcl.Forms, Vcl.Dialogs, Vcl.StdCtrls, Zip;

...

procedure TForm1.Button1Click(Sender: TObject);
var
success: Boolean;
zip: HCkZip;
xml: PWideChar;

begin
success := False;

success := False;

// Open an existing ZIP archive.
zip := CkZip_Create();

success := CkZip_OpenZip(zip,'example.zip');
if (success = False) then
  begin
    Memo1.Lines.Add(CkZip__lastErrorText(zip));
    Exit;
  end;

// Get the ZIP directory information as XML.
xml := CkZip__getDirectoryAsXML(zip);

Memo1.Lines.Add(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>

CkZip_CloseZip(zip);

Memo1.Lines.Add('Done.');

CkZip_Dispose(zip);

end;