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Java.nio.file.path for a classpath resource

java
java-8
path
file-system
Alex KataevbyAlex Kataev·Nov 14, 2024
TLDR

To swiftly convert a classpath resource to a Path, utilize ClassLoader.getResource and Paths.get provided the resource is accessible via the file system.

URL resource = Thread.currentThread().getContextClassLoader().getResource("resource.txt"); Path path = resource != null ? Paths.get(resource.toURI()) : null;

This method works for file-based resources, not within JARs. For resources inside JARs, consider first extracting to a temporary file.

Converting JARs resources to Path

With Files.copy, shift the resource to a temporary file and utilize it as a Path. This technique is imperative to evade FileSystemNotFoundException when resource is within a JAR file and can't be directly transformed to Path.

InputStream resourceStream = classLoader.getResourceAsStream("resource.txt"); Path tempPath = Files.createTempFile(null, ".txt"); // Here, creating a temporary file. Files.copy(resourceStream, tempPath, StandardCopyOption.REPLACE_EXISTING); // Voilà! Your resource.txt from JAR is now available as a Path.

While handling the conversion of a URL to a URI, one common stumbling block is the URISyntaxException. Demonstrate due diligence in error handling.

try { Path path = Paths.get(MyClass.class.getResource("resource.txt").toURI()); } catch (URISyntaxException e) { System.err.println("Oops, URL to URI conversion took a (wrong) turn!"); // Even an error message can crack a smile! }

Getting resource-savvy with URI schemes

To correctly handle resource locations, it's crucial to validate the URI scheme accounting for both "file" and "jar" types.

String scheme = resource.getProtocol(); if (!"file".equals(scheme) && !"jar".equals(scheme)) { throw new IllegalArgumentException("Unsupported scheme: " + scheme); // Here be dragons 🐉 }

Streamlining with Lambda

In Java 8 and beyond, your code can be more concise when handling classpath resources. Employ lambda expressions or method references.

Files.lines(Path.of(classLoader.getResource("config.properties").toURI())) .forEach(System.out::println); // Line-by-line. Efficient, almost poetic!

Safe handling of resource streams

Implement try-with-resources to ensure streams are safely closed, avoiding resource leaks.

try (BufferedReader reader = new BufferedReader( new InputStreamReader(classLoader.getResourceAsStream("resource.txt")))) { reader.lines().forEach(System.out::println); // Every stream has a safe ending. :) }

This is of utmost importance, as it prevents file handle exhaustion.

Path conversion of JAR resources

For resources within JAR files, leverage a custom filesystem via the Zip File System Provider for Path transformations.

Map<String, String> env = new HashMap<>(); env.put("create", "true"); URI jarURI = URI.create("jar:file:/path/to/your.jar"); try (FileSystem fs = FileSystems.newFileSystem(jarURI, env)) { Path pathInJar = fs.getPath("/resource-in-jar.txt"); // Navigate the JAR like it's a local file system. } // Bid farewell to the FileSystem.

Keeping your code compatible

Strive to ensure compatibility across Java versions. Your code should be version agnostic, running without hiccups from Java 7 through Java 8 and up to Java 18.

Special handling for classpath resources

As classpath resources come with nuances, some environments require tailoring. For instance, when operating in an environment with a security manager, use AccessController.doPrivileged to eschew SecurityException.