How Can I List Files Inside a JAR File in Java?

Listing Files Within a JAR File

In Java, the listFiles() method can be used to obtain a list of files within a directory. However, it cannot directly be applied to a JAR file to retrieve its contents.

To achieve this, you can utilize Java's CodeSource class:

CodeSource src = MyClass.class.getProtectionDomain().getCodeSource();
if (src != null) {
  URL jar = src.getLocation();
  try (ZipInputStream zip = new ZipInputStream(jar.openStream())) {
    while (true) {
      ZipEntry e = zip.getNextEntry();
      if (e == null) {
        break;
      }
      String name = e.getName();
      if (name.startsWith("path/to/your/dir/")) {
        // Process the entries within the specified directory.
      }
    }
  } catch (IOException ex) {
    // Handle I/O exceptions.
  }
} else {
  // Handle the case where no CodeSource is available.
}

Note:

• The ZipInputStream reads the JAR file as a ZIP archive.
• The ZipEntry class represents individual entries (files) within the JAR file.
• The getNextEntry() method retrieves the next entry in the JAR file, allowing you to iterate through its contents.

Java 7 Enhancements:

In Java 7 and later, you can create a FileSystem object directly from the JAR file, using the FileSystems.newFileSystem() method. This allows you to access JAR file contents using NIO's directory walking and filtering mechanisms, making it more convenient to handle both JARs and "exploded" directories.

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