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Articles > Misc > How to create Zip stream/file content in-memory

How to create Zip stream/file content in-memory

Author: Venkata Sudhakar

The below example shows how to create a Zip stream/file content in-memory.

package com.bethecoder.utils;

import java.io.ByteArrayOutputStream;
import java.io.File;
import java.util.zip.ZipEntry;
import java.util.zip.ZipOutputStream;

import org.apache.commons.io.FileUtils;

public class ZipDirectory {

public static void main(String[] args) {
		byte [] zipContent = createZipByteArray(new File("C:\\DIR_TO_ZIP"));
	}

public static byte[] createZipByteArray(File directoryToZip) {
		byte[] zipContent = new byte[0];
		try {
			ByteArrayOutputStream output = new ByteArrayOutputStream();
	        ZipOutputStream zipOut = new ZipOutputStream(output);
	        for (File fileToZip : directoryToZip.listFiles()) {
	        	if (fileToZip.isFile()) {
		            ZipEntry zipEntry = new ZipEntry(fileToZip.getName());
		            zipOut.putNextEntry(zipEntry);
		            zipOut.write(FileUtils.readFileToByteArray(fileToZip));
	        	}
	        }
	        zipOut.close();
	        zipContent = output.toByteArray();
		} catch (Exception e) {
			e.printStackTrace();
		}
		return zipContent;
    }
}

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