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About Byron Kiourtzoglou

Byron Kiourtzoglou
Byron is a master software engineer working in the IT and Telecom domains. He is an applications developer in a wide variety of applications/services. He is currently acting as the team leader and technical architect for a proprietary service creation and integration platform for both the IT and Telecom industries in addition to a in-house big data real-time analytics solution. He is always fascinated by SOA, middleware services and mobile development. Byron is co-founder and Executive Editor at Java Code Geeks.

Create zip file from multiple files with ZipOutputStream

With this example we are going to demonstrate how to create a zip file from multiple Files with ZipOutputStream, that implements an output stream filter for writing files in the ZIP file format. In short, to create a zip file from multiple Files with ZipOutputStream you should:

  • Create a FileOutputStream to write to the file with the specified name, that is the zipFile.
  • Create a new ZipOutputStream from the FileOutputStream, that is an output stream filter for writing files in the ZIP file format.
  • For each one of the Files create a new File instance by the given pathname of the file.
  • Create a FileInputStream by opening a connection to the file.
  • Create a new ZipEntry with the name of the File and begin writing it to the ZipOutputStream. The default compression method will be used if no compression method was specified for the entry.
  • Read up to 1024 bytes of data from the file into an array of bytes, using the read(byte[] b) API method of FileInputStream and write the data to the current ZipEntry data, using write(byte[] b, int off, int len) method of ZipOutputStream.
  • Always remember to close the ZipEntry, the ZipOutputStream and the FileInputStream, using closeEntry() and close() API methods respectively.

Let’s take a look at the code snippet that follows:

package com.javacodegeeks.snippets.core;

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;

public class CreateZipFileFromMultipleFilesWithZipOutputStream {
	
	public static void main(String[] args) {
		
		String zipFile = "C:/archive.zip";
		
		String[] srcFiles = { "C:/srcfile1.txt", "C:/srcfile2.txt", "C:/srcfile3.txt"};
		
		try {
			
			// create byte buffer
			byte[] buffer = new byte[1024];

			FileOutputStream fos = new FileOutputStream(zipFile);

			ZipOutputStream zos = new ZipOutputStream(fos);
			
			for (int i=0; i < srcFiles.length; i++) {
				
				File srcFile = new File(srcFiles[i]);

				FileInputStream fis = new FileInputStream(srcFile);

				// begin writing a new ZIP entry, positions the stream to the start of the entry data
				zos.putNextEntry(new ZipEntry(srcFile.getName()));
				
				int length;

				while ((length = fis.read(buffer)) > 0) {
					zos.write(buffer, 0, length);
				}

				zos.closeEntry();

				// close the InputStream
				fis.close();
				
			}

			// close the ZipOutputStream
			zos.close();
			
		}
		catch (IOException ioe) {
			System.out.println("Error creating zip file: " + ioe);
		}
		
	}

}

 
This was an example of how to create a zip file from multiple Files with ZipOutputStream in Java.

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