Classpath resource not found when running as jar

java

resource-management

exception-handling

best-practices

byAlex Kataev·Dec 26, 2024

Avoiding a missing classpath resource in a .jar file is possible by leveraging getClass().getResourceAsStream("/path/to/resource"). The leading slash makes a difference - it informs getResourceAsStream to initiate from the classpath root:

InputStream resourceStream = getClass().getResourceAsStream("/path/to/resource");
if (resourceStream == null) {
    throw new RuntimeException("Resource not found!");  // Because shouting helps.😉
}

Ensure the path is legitimate, and refrain from utilizing File or FileInputStream, they'll be of no use in .jar.

The getFile() Radius is Restricted

When your app runs from an IDE, for example, STS (Spring Tool Suite), it might access classpath resources employing getFile(). This approach fails in executable JAR considering getFile() tries reaching the file system and can't handle resources nested in jar files, throwing java.io.FileNotFoundException.

Resource Access Best Practices

ClassPathResource is your friend when dealing with resources in Spring Boot. It offers an efficient and accessible way for classpath resources.
To read the content as bytes, shake hands with FileCopyUtils.copyToByteArray().
When converting an InputStream to a string, be specific about your Charset, for example, StandardCharsets.UTF_8.
Always do a reality check for resource inclusion during the build and ensure its location is consistent inside the jar.

Strong Exception Handling

Welcome IOException with open arms using try-catch blocks. This ensures the show goes on even in the face of exceptions and closely guards resources against code leaks. If checked exceptions aren't your cup of tea, wrap an IOException inside an UncheckedIOException for a smoother exception flow.

Effective Resource Management for Everyone

Resource Reading with Characters

When it's literary assets you're dealing with, BufferedReader along with InputStreamReader is a great approach:

InputStream is = getClass().getResourceAsStream("/path/to/resource");
BufferedReader reader = new BufferedReader(new InputStreamReader(is, StandardCharsets.UTF_8));

Trust me, this approach loves to read the resource line by line or in your preferred patterns.

Dynamic is the New Static

Stay away from hardcoding resource paths. Take control by defining locations in properties files or flex your dynamism with setter methods. Here's how you can list all *.json files in a resource folder like a boss:

ResourcePatternResolver resolver = ResourcePatternUtils.getResourcePatternResolver(new DefaultResourceLoader());
Resource[] resources = resolver.getResources("classpath:" + location + "/*.json");