Java examples to read a file from the resources folder in either a simple Java application or a Spring MVC / Boot application.
Table of Contents 1. Setup 2. ClassLoader.getResource() 3. ResourceUtils.getFile()
1. Setup
Below image describes the folder structure used in this example. Notice the file sample.txt is in /src/main/resources folder.
2. ClassLoader getResource() and getResourceAsStream()
Methods in the classes Class and ClassLoader provide a location-independent way to locate resources. We can read a file from the application’s resources package by using ClassLoader reference.
The method getResource() returns a URL for the resource. If the resource does not exist or is not visible due to security considerations, the methods return null.
The getResource() and getResourceAsStream() methods find a resource with a given name. They return null if they do not find a resource with the specified name.
getResourceAsStream()returns anInputStreamfor the resource.getResource()returns a URL for the resource.
Example 1: Java program to read a file from resources folder using getResource() method
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 = 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);
}
}
Program output:
File Found : true Test Content
Example 2: Java program to read a file from resources folder using getResourceAsStream() method
package com.howtodoinjava.demo;
import java.io.File;
import java.io.IOException;
import java.nio.file.Files;
import org.apache.commons.io.IOUtils;
public class ReadResourceFileDemo
{
public static void main(String[] args) throws IOException
{
String fileName = "config/sample.txt";
ClassLoader classLoader = getClass().getClassLoader();
try (InputStream inputStream = classLoader.getResourceAsStream(fileName)) {
String result = IOUtils.toString(inputStream, StandardCharsets.UTF_8);
System.out.println(result);
} catch (IOException e) {
e.printStackTrace();
}
}
}
Program output:
Test Content
3. ResourceUtils.getFile()
If your application happens to be Spring WebMVC or Spring Boot application then you may directly take advantage of ResourceUtils class.
Example 3: Java program to read a file from resources folder using ResourceUtils
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:
File Found : true Test Content
Happy Learning !!
Hien Nguyen
Thank you so much for sharing. This helps me a lot
griffin
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())); } }ARM
Solve my problem, thanks!
Aftab
will this procedure work for a .xlsx file too?
Charl
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
Lokesh Gupta
Thanks for sharing. Much appreciated !!
sandeep
Thanks it works !!!
Saul
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”
Vikas
Thanks it works
Vikas