Java Language Tutorial => Accessing the contents of a ZIP file

Example

The FileSystem API of Java 7 allows to read and add entries from or to a Zip file using the Java NIO file API in the same way as operating on any other filesystem.

The FileSystem is a resource that should be properly closed after use, therefore the try-with-resources block should be used.

Reading from an existing file

Path pathToZip = Paths.get("path/to/file.zip");
try(FileSystem zipFs = FileSystems.newFileSystem(pathToZip, null)) {
  Path root = zipFs.getPath("/");
  ... //access the content of the zip file same as ordinary files
} catch(IOException ex) {
  ex.printStackTrace();
}

Creating a new file

Map<String, String> env = new HashMap<>();  
env.put("create", "true"); //required for creating a new zip file
env.put("encoding", "UTF-8"); //optional: default is UTF-8
URI uri = URI.create("jar:file:/path/to/file.zip");
try (FileSystem zipfs = FileSystems.newFileSystem(uri, env)) {
  Path newFile = zipFs.getPath("/newFile.txt");
  //writing to file
  Files.write(newFile, "Hello world".getBytes());
} catch(IOException ex) {
  ex.printStackTrace();
}

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Java Language

Fastest Entity Framework Extensions

Example

Reading from an existing file

Creating a new file

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Java Language

Java Language File I/O Accessing the contents of a ZIP file

Fastest Entity Framework Extensions

Example

Reading from an existing file

Creating a new file

Got any Java Language Question?