What is the best way to extract a zip file using java

Question

I have an archived file. This file also contains various directories and files. I want to extract all of these and save on the specified path.

So how to write a java program to extract the archived file.

Thanks Sunil Kumar Sahoo

Answer 1

public static  void unzip(){
      try{
           BufferedOutputStream out = null;
           ZipInputStream  in = new ZipInputStream(ZipFileExtracter.class.getClassLoader().getResourceAsStream("com/artificialmachines/chitme/stamps/ChitMeData.zip"));
           ZipEntry entry;
           boolean isDirectory=false;
           while((entry = in.getNextEntry()) != null){
               int count;
               byte data[] = new byte[BUFFER];
               // write the files to the disk
               String entryName = entry.getName();
               File newFile = new File(new StringBuffer().append(System.getProperty("user.dir")).append(File.separator).append(entryName).toString());
               if(entryName.endsWith("/")){
                   isDirectory=true;
                   newFile.mkdir();
                   //System.out.println("This is directory "+newFile.exists()+"  IS DIr "+newFile.isDirectory()+"    path "+newFile.getPath());
               }else{
                   newFile.createNewFile();
               }
               if(!isDirectory){
                   out = new BufferedOutputStream(new FileOutputStream(newFile),BUFFER);
                    while ((count = in.read(data,0,BUFFER)) != -1){
                         out.write(data,0,count);
                    }
                    cleanUp(out);
               }
               isDirectory=false;
           }
           cleanUp(in);
      }
      catch(Exception e){
           e.printStackTrace();
           System.exit(0);
      }
   }

Answer 2

You must study the java.util.zippackage . Sun provides a tutorial on its use: Compressing and decompressing data using the Java API .

Answer 3

java.util.zip. * . API refernec. LookThis