Dart

Unzip Files Matching a Pattern (such as *.xml)

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Example code to unzip Files Matching a Pattern (such as *.xml) Dart

import 'package:chilkat/chilkat.dart';

void main() {
  // This example works with the .zip located at http://www.example-code.com/data/test.zip

  // This example assumes the Chilkat API to have been previously unlocked.
  // See Global Unlock Sample for sample code.

  // First, let's download the .zip
  final http = CkHttp();
  try {
    http.download('http://www.example-code.com/data/test.zip', 'qa_test/test.zip');
  } on ChilkatException catch (e) {
    print(e.lastErrorText);
    return;
  }

  final zip = CkZip();

  try {
    zip.openZip('qa_test/test.zip');
  } on ChilkatException catch (e) {
    print(e.lastErrorText);
    return;
  }

  // Show what's in the .zip by getting the directory as XML.
  final String zipContentsXml;
  try {
    zipContentsXml = zip.getDirectoryAsXML();
  } on ChilkatException catch (e) {
    print(e.lastErrorText);
    return;
  }

  print(zipContentsXml);

  // In this case, the contents of our .zip look like this:
  // <?xml version="1.0" encoding="utf-8" ?>
  // <zip_contents>
  //     <dir name="test">
  //         <dir name="abc">
  //             <dir name="123">
  //                 <file>ecc_public.pem</file>
  //                 <file>starfish.jpg</file>
  //             </dir>
  //             <file>comodo-certificates.pem</file>
  //             <file>pigs.xml</file>
  //         </dir>
  //         <file>COMODORSACertificationAuthority.crt</file>
  //         <file>hamlet.xml</file>
  //     </dir>
  // </zip_contents>

  // ------------------------------------------------------------------------
  // If we wish to unzip only .xml files, use "*.xml" for the pattern\
  // Unzip to the "qa_output" directory relative to our current working directory.
  // (You may unzip to an absolute path by passing an absolute path, such as "/someDir" or
  // "C:/someDir", etc.)
  // This will unzip only the .xml files.  The relative directory structure within
  // the .zip will be created in the local filesystem under the unzip directory
  // path passed in the first argument, which in this case is "qa_output".
  var count = zip.unzipMatching('qa_output', '*.xml', true);
  if (count < 0) {
    print(zip.lastErrorText);
    return;
  }

  print('unzipped $count matching file(s).');

  // ------------------------------------------------------------------------
  // If we wish to unzip only files under the "123" directory, we can do this:
  count = zip.unzipMatching('qa_output2', '*/123/*', true);
  if (count < 0) {
    print(zip.lastErrorText);
    return;
  }

  print('unzipped $count matching file(s).');

  // ------------------------------------------------------------------------
  // If no asterisk character is used, then the match must be exact.
  // For example, to unzip "test/hamlet.xml" we must pass the exact path
  // as it is in the .zip.  We cannot just pass "hamlet.xml".
  count = zip.unzipMatching('qa_output3', 'test/hamlet.xml', true);
  if (count < 0) {
    print(zip.lastErrorText);
    return;
  }

  print('unzipped $count matching file(s).');

  // ------------------------------------------------------------------------
  // If we want to unzip any file named "hamlet.xml" found in any sub-directory...
  count = zip.unzipMatching('qa_output4', '*/hamlet.xml', true);
  if (count < 0) {
    print(zip.lastErrorText);
    return;
  }

  print('unzipped $count matching file(s).');
}