HowToDoInJava

Read file from resources folder – Spring Example

By | Filed Under: Java I/O

Two Java examples of reading a file from resources folder in either simple Java application or any spring mvc / spring boot application.

Table of Contents

Project Structure
Read file using ClassLoader.getResource().toURI()
Read file using Spring's ResourceUtils.getFile()

Project Structure

Below image describes the folder structure used in this example. Notice the file sample.txt in resources folder.

Read file from resources folder
Read file from resources folder

Read file from resources folder using ClassLoader.getResource().toURI()

We can read a file from the application’s resources package by using ClassLoader reference of instance of the class.

package com.howtodoinjava.demo;

import java.io.File;
import java.io.IOException;
import java.nio.file.Files;

public class ReadResourceFileDemo 
{
	public static void main(String[] args) throws IOException 
	{
		String fileName = "config/sample.txt";
		ClassLoader classLoader = new ReadResourceFileDemo().getClass().getClassLoader();

		File file = new File(classLoader.getResource(fileName).getFile());
		
		//File is found
		System.out.println("File Found : " + file.exists());
		
		//Read File Content
		String content = new String(Files.readAllBytes(file.toPath()));
		System.out.println(content);
	}
}

We can avoid creating a un-necessary instance of the class if we utilize system classloader instance as below-

String fileName = "config/sample.txt";

ClassLoader classLoader = ClassLoader.getSystemClassLoader();

File file = new File(classLoader.getResource(fileName).getFile());

//File is found
System.out.println("File Found : " + file.exists());

//Read File Content
String content = new String(Files.readAllBytes(file.toPath()));
System.out.println(content);

Program output is shown below.

Output:

File Found : true
Test Content

Read file in Spring – ResourceUtils.getFile()

If your application happens to be spring or spring boot based application then you may directly take advantage of ResourceUtils class.

File file = ResourceUtils.getFile("classpath:config/sample.txt")

//File is found
System.out.println("File Found : " + file.exists());

//Read File Content
String content = new String(Files.readAllBytes(file.toPath()));
System.out.println(content);

Program output is shown below.

Output:

File Found : true
Test Content

Happy Learning !!

References:

Spring ResourceUtils
ClassLoader.getResource()

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Feedback, Discussion and Comments

  1. Works either from filesystem and JAR:

    	public static void main(String[] args) throws IOException {

		String resourceFile = "resource/test_resource.txt";

		InputStream resourceStream = ClassLoader.getSystemClassLoader().getResourceAsStream(resourceFile);

		if (resourceStream != null) {
			BufferedReader resReader = new BufferedReader(new InputStreamReader(resourceStream));
			System.out.println(resReader.lines().collect(Collectors.joining()));
		}
	}
    Reply

  2. Solve my problem, thanks!

    Reply

  3. will this procedure work for a .xlsx file too?

    Reply

  4. The class loader method works when I run it in IDE, but not when the application is packaged as a jar file. I had to use a ZIP file system to load it from a jar file e.g

    	private byte[] loadData() throws Exception {

		String relativePath = "META-INF/somebinaryfile.bmp";
		URI resource = getClass().getClassLoader().getResource(relativePath).toURI();

		byte[] data = null;
		try(FileSystem jarFileSystem = FileSystems.getFileSystem(resource)) {
			data = readDataFromFile(relativePath, jarFileSystem);
		} catch (FileSystemNotFoundException e) {
			 Map<String, String> env = new HashMap<String, String>();
			 env.put("create", "true");
			try (FileSystem jarFileSystem = FileSystems.newFileSystem(resource, env)) {
				data = readDataFromFile(relativePath, jarFileSystem);
			}
		}
		return data;
	}

	private byte[] readDataFromFile(String fileName, FileSystem jarFileSystem) throws IOException {

		Path path = jarFileSystem.getPath(fileName);
		byte[] data = Files.readAllBytes(path);
		return data;
	}

    Also see this: https://docs.oracle.com/javase/7/docs/technotes/guides/io/fsp/zipfilesystemprovider.html

    Reply

  5. Thanks it works !!!

    Reply

  6. not working for me, I’m getting this error “Exception in thread “main” java.lang.NullPointerException” and if I try using the filename without folder the error is “\target\classes\sample.txt”

    Reply

  7. Thanks it works

    Vikas

    Reply

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