How to Extract zip File from jar file

Question

I am having a zip file into my project. When I am running my code through IDE, my extract(String file, String destination) method works fine.

 D:/Tools/JAVA/Lodable_Creation/build/classes/ib2.zip-->
 String s1=getClass().getResource("Test.zip").getPath().toString();
  extract(s1, "c:\\");

This is giving me Path s1 is--> D:\\Tools\\JAVA\\Lodable_Creation\\build

When I compile same code and run through Command prompt

file:/D:/Tools/JAVA/Lodable_Creation/dist/Lodable_Creation.jar!/Test.zip
s1 is-->D:\Tools\JAVA\Lodable_Creation\dist

And I am not getting output. Please help me.

UPDATE:-

public static void extract(String file, String destination) throws IOException {
    ZipInputStream in = null;
    OutputStream out = null;
    try {
      // Open the ZIP file
      in = new ZipInputStream(new FileInputStream(file));
      // Get the first entry
      ZipEntry entry = null;
      while ((entry = in.getNextEntry()) != null) {
        String outFilename = entry.getName();
        // Open the output file
        if (entry.isDirectory()) {
          new File(destination, outFilename).mkdirs();
        } else {
          out = new FileOutputStream(new File(destination,outFilename));
          // Transfer bytes from the ZIP file to the output file
          byte[] buf = new byte[1024];
          int len;
          while ((len = in.read(buf)) > 0) {
            out.write(buf, 0, len);
          }
          out.close();
        }
      }
    } finally {
      // Close the stream
      if (in != null) {
        in.close();
      }
      if (out != null) {
        out.close();
      }
    }
  }

On Ok Click button

Map map = System.getenv();
 Set keys = map.keySet();
 String newString  = (String) map.get("CYGWIN_HOME");
 System.out.println(" " + newString);
 String  destination= newString.replace(";", "");
 System.out.println(" " + destination);
 String S =getClass().getResource("Test.zip").getPath().toString();
 File jarFile = new File(S);
 String file=jarFile.toString();
 extract(file,destination);

This is my actual code for extract method and on OK Button . This is extracting Test.zip file to the Destination folder. ie CYGWIN_HOME

Answer 1

如果您的文件路径实际上是URL（以"file://"开头），则使用new ZipInputStream((new URL(file)).openStream())否则使用new ZipInputStream(new FileInputStream(file))就像您已经做过的一样。