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 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()
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