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| Articles > Misc > How to create Zip stream/file content in-memory | |
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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.
01 | package com.bethecoder.utils; |
03 | import java.io.ByteArrayOutputStream; |
05 | import java.util.zip.ZipEntry; |
06 | import java.util.zip.ZipOutputStream; |
08 | import org.apache.commons.io.FileUtils; |
10 | public class ZipDirectory { |
12 | public static void main(String[] args) { |
13 | byte [] zipContent = createZipByteArray(new File("C:\\DIR_TO_ZIP")); |
16 | public static byte[] createZipByteArray(File directoryToZip) { |
17 | byte[] zipContent = new byte[0]; |
19 | ByteArrayOutputStream output = new ByteArrayOutputStream(); |
20 | ZipOutputStream zipOut = new ZipOutputStream(output); |
21 | for (File fileToZip : directoryToZip.listFiles()) { |
22 | if (fileToZip.isFile()) { |
23 | ZipEntry zipEntry = new ZipEntry(fileToZip.getName()); |
24 | zipOut.putNextEntry(zipEntry); |
25 | zipOut.write(FileUtils.readFileToByteArray(fileToZip)); |
29 | zipContent = output.toByteArray(); |
30 | } catch (Exception e) { |
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