How to Create and Extract a ZIP File in Java? Unzip Example Tutorial

In Java, you can use ZipFile class to unzip a .jar and .zip file in Java. These classes are defined in and java.util.zip package. Unfortunately there is no ZipUtil class, which provides straightforward methods to decompress zip files, but thankfully there are enough tools available in JDK to write your own method to extract zip file in Java. There are mainly two steps to unzip a zip file in Java, first, create all directories inside of the zip file, because zip archive flattens all directory and second create all file which are inside .zip file. Sinze files in .zip file is not in specific order, you need to iterate twice to perform these steps individually.

Just ensure not to read and write content byte by byte While extracting zip file, because it's not efficient and result in poor performance while decompressing large ZIP or gz file. You can also create Java programs to create a zip file by using java.util.zip package.

How to unzip file to directory in Java?

To unzip a file to a specific directory in Java, you can use the java.util.zip package to read the ZIP file and extract its contents. 

Here's a step-by-step guide on how to do it in Java program:

1. Import the necessary classes

First, import the required classes from the java.util.zip and java.io packages:

import java.io.File;

import java.io.FileOutputStream;

import java.io.IOException;

import java.io.InputStream;

import java.util.Enumeration;

import java.util.zip.ZipEntry;

import java.util.zip.ZipFile;

2. Create a method to unzip the file

You can create a method that takes the path to the ZIP file and the target directory where you want to extract the contents. 

Here's an example method

public static void unzipFileToDirectory(String zipFilePath, 
    String targetDirectory) throws IOException {

    File directory = new File(targetDirectory);



    // Create the target directory if it doesn't exist

    if (!directory.exists()) {

        directory.mkdirs();

    }



    try (ZipFile zipFile = new ZipFile(zipFilePath)) {

        Enumeration<? extends ZipEntry> entries = zipFile.entries();



        while (entries.hasMoreElements()) {

            ZipEntry entry = entries.nextElement();

            String entryName = entry.getName();

            File entryFile = new File(directory, entryName);



            // Create intermediate directories if they don't exist

            if (entry.isDirectory()) {

                entryFile.mkdirs();

            } else {

                // Create the parent directory for the file

                File parent = entryFile.getParentFile();

                if (!parent.exists()) {

                    parent.mkdirs();

                }



                // Extract the file

                try (InputStream inputStream = zipFile.getInputStream(entry);

                     FileOutputStream outputStream = new FileOutputStream(entryFile)) {

                    byte[] buffer = new byte[1024];

                    int bytesRead;

                    while ((bytesRead = inputStream.read(buffer)) != -1) {

                        outputStream.write(buffer, 0, bytesRead);

                    }

                }

            }

        }

    }

}

3. Call the method to unzip the file

Now, you can call the unzipFileToDirectory() method with the path to the ZIP file and the target directory where you want to extract the contents. For example:

public static void main(String[] args) {

    String zipFilePath = "path/to/your/file.zip";

    String targetDirectory = "path/to/your/target/directory";



    try {

        unzipFileToDirectory(zipFilePath, targetDirectory);

        System.out.println("File unzipped successfully.");

    } catch (IOException e) {

        e.printStackTrace();

    }

}

Just replace "path/to/your/file.zip" with the actual path to your ZIP file and "path/to/your/target/directory" with the directory where you want to extract the contents.

If you prefer a more streamlined approach to extracting files from ZIP archives, you can use third-party libraries like Zip4j. Zip4j simplifies the process of extracting all files from a ZIP archive with ease, making it a handy tool for your Java projects.

In summary, unzipping ZIP files in Java involves recreating directories and extracting files from the archive. Be sure to avoid inefficient byte-by-byte copying and use buffered streams for better performance. 

Additionally, consider using third-party libraries like Zip4j to simplify the extraction process. Keep these tips in mind, and you'll be well-equipped to handle ZIP files in your Java applications.

How to create a Zip file in Java?

To create a ZIP file in Java, you can use the java.util.zip package. Here's a step-by-step guide on how to create a ZIP file:

1. Import the necessary classes

First, import the required classes from the java.util.zip and java.io packages:

import java.io.File;

import java.io.FileInputStream;

import java.io.FileOutputStream;

import java.io.IOException;

import java.util.zip.ZipEntry;

import java.util.zip.ZipOutputStream;

2. Create a method to create the ZIP file

You can create a method that takes a list of files or directories that you want to include in the ZIP file and the name of the ZIP file to create. Here's an example method:

public static void createZipFile(String zipFileName, String... filesToAdd)
  throws IOException {

    try (FileOutputStream fos = new FileOutputStream(zipFileName);

         ZipOutputStream zos = new ZipOutputStream(fos)) {



        for (String fileToAdd : filesToAdd) {

            File file = new File(fileToAdd);

            if (file.exists()) {

                addFileToZip(zos, file, "");

            }

        }

    }

}

private static void addFileToZip(ZipOutputStream zos, File file, String parentDirectory) throws IOException {

    String entryName = parentDirectory + file.getName();



    if (file.isDirectory()) {

        File[] files = file.listFiles();

        if (files != null) {

            for (File nestedFile : files) {

                addFileToZip(zos, nestedFile, entryName + "/");

            }

        }

    } else {

        try (FileInputStream fis = new FileInputStream(file)) {

            ZipEntry zipEntry = new ZipEntry(entryName);

            zos.putNextEntry(zipEntry);



            byte[] buffer = new byte[1024];

            int bytesRead;

            while ((bytesRead = fis.read(buffer)) != -1) {

                zos.write(buffer, 0, bytesRead);

            }

        }

    }

}

The createZipFile() method takes the name of the ZIP file to create and an array of file or directory paths to include in the ZIP file. It uses the addFileToZip() method to add each file or directory to the ZIP file.

3. Call the method to create the ZIP file

You can call the createZipFile method to create the ZIP file with the desired contents. For example:

public static void main(String[] args) {

    String zipFileName = "path/to/your/output.zip";

    String[] filesToAdd = {

        "path/to/your/file1.txt",

        "path/to/your/directory1",

        "path/to/your/file2.txt"

    };



    try {

        createZipFile(zipFileName, filesToAdd);

        System.out.println("ZIP file created successfully.");

    } catch (IOException e) {

        e.printStackTrace();

    }

}

Just don't forget to replace "path/to/your/output.zip" with the desired path and name for the output ZIP file, and specify the files or directories you want to include in the filesToAdd array. That's it! This code will create a ZIP file containing the specified files and directories.

That's all about how to unzip a ZIP file in Java. You can also use these steps to extract content of .gz file in UNIX because .gz is nothing but .zip file. Keep in mind, not to copy data byte by byte because its not efficient, use buffer to buffer copy for better performance. There are open source library also available for dealing with zip file in Java.