How do I load a file from resource folder?

My project has the following structure:

/src/main/java/
/src/main/resources/
/src/test/java/
/src/test/resources/

I have a file in /src/test/resources/test.csv and I want to load the file from a unit test in /src/test/java/MyTest.java

I have this code which didn't work. It complains "No such file or directory".

BufferedReader br = new BufferedReader (new FileReader(test.csv))

I also tried this

InputStream is = (InputStream) MyTest.class.getResourcesAsStream(test.csv))

This also doesn't work. It returns null. I am using Maven to build my project.

Try the next:

ClassLoader classloader = Thread.currentThread().getContextClassLoader();
InputStream is = classloader.getResourceAsStream("test.csv");

---

If the above doesn't work, various projects have been added the following class: ClassLoaderUtil¹ (code here).²

Here are some examples of how that class is used:

src\main\java\com\company\test\YourCallingClass.java
src\main\java\com\opensymphony\xwork2\util\ClassLoaderUtil.java
src\main\resources\test.csv

// java.net.URL
URL url = ClassLoaderUtil.getResource("test.csv", YourCallingClass.class);
Path path = Paths.get(url.toURI());
List<String> lines = Files.readAllLines(path, StandardCharsets.UTF_8);

// java.io.InputStream
InputStream inputStream = ClassLoaderUtil.getResourceAsStream("test.csv", YourCallingClass.class);
InputStreamReader streamReader = new InputStreamReader(inputStream, StandardCharsets.UTF_8);
BufferedReader reader = new BufferedReader(streamReader);
for (String line; (line = reader.readLine()) != null;) {
	// Process line
}

---

Notes

1. See it in The Wayback Machine.
2. Also in GitHub.

Try:

InputStream is = MyTest.class.getResourceAsStream("/test.csv");

IIRC getResourceAsStream() by default is relative to the class's package.

As @Terran noted, don't forget to add the / at the starting of the filename

Try following codes on Spring project

ClassPathResource resource = new ClassPathResource("fileName");
InputStream inputStream = resource.getInputStream();

Or on non spring project

 ClassLoader classLoader = getClass().getClassLoader();
 File file = new File(classLoader.getResource("fileName").getFile());
 InputStream inputStream = new FileInputStream(file);

Here is one quick solution with the use of Guava:

import com.google.common.base.Charsets;
import com.google.common.io.Resources;

public String readResource(final String fileName, Charset charset) throws IOException {
        return Resources.toString(Resources.getResource(fileName), charset);
}

Usage:

String fixture = this.readResource("filename.txt", Charsets.UTF_8)

Non spring project:

String filePath = Objects.requireNonNull(getClass().getClassLoader().getResource("any.json")).getPath();

Stream<String> lines = Files.lines(Paths.get(filePath));

Or

String filePath = Objects.requireNonNull(getClass().getClassLoader().getResource("any.json")).getPath();

InputStream in = new FileInputStream(filePath);

For spring projects, you can also use one line code to get any file under resources folder:

File file = ResourceUtils.getFile(ResourceUtils.CLASSPATH_URL_PREFIX + "any.json");

String content = new String(Files.readAllBytes(file.toPath()));

For java after 1.7

 List<String> lines = Files.readAllLines(Paths.get(getClass().getResource("test.csv").toURI()));

Alternatively you can use Spring utils if you are in Spring echosystem

final val file = ResourceUtils.getFile("classpath:json/abcd.json");

To get to more behind the scene, check out following blog

https://todzhang.com/blogs/tech/en/save_resources_to_files

I faced the same issue.

The file was not found by a class loader, which means it was not packed into the artifact (jar). You need to build the project. For example, with maven:

mvn clean package

So the files you added to resources folder will get into maven build and become available to the application.

I would like to keep my answer: it does not explain how to read a file (other answers do explain that), it answers why InputStream or resource was null. Similar answer is here.

ClassLoader loader = Thread.currentThread().getContextClassLoader();
InputStream is = loader.getResourceAsStream("test.csv");

If you use context ClassLoader to find a resource then definitely it will cost application performance.

Now I am illustrating the source code for reading a font from maven created resources directory,

> scr/main/resources/calibril.ttf

Font getCalibriLightFont(int fontSize){
	Font font = null;
	try{
		URL fontURL = OneMethod.class.getResource("/calibril.ttf");
		InputStream fontStream = fontURL.openStream();
		font = new Font(Font.createFont(Font.TRUETYPE_FONT, fontStream).getFamily(), Font.PLAIN, fontSize);
		fontStream.close();
	}catch(IOException | FontFormatException ief){
		font = new Font("Arial", Font.PLAIN, fontSize);
		ief.printStackTrace();
	}	
	return font;
}

It worked for me and hope that the entire source code will also help you, Enjoy!

Import the following:

import java.io.IOException;
import java.io.FileNotFoundException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
import java.util.ArrayList;

The following method returns a file in an ArrayList of Strings:

public ArrayList<String> loadFile(String filename){

  ArrayList<String> lines = new ArrayList<String>();

  try{

    ClassLoader classloader = Thread.currentThread().getContextClassLoader();
    InputStream inputStream = classloader.getResourceAsStream(filename);
    InputStreamReader streamReader = new InputStreamReader(inputStream, StandardCharsets.UTF_8);
    BufferedReader reader = new BufferedReader(streamReader);
    for (String line; (line = reader.readLine()) != null;) {
      lines.add(line);
    }

  }catch(FileNotFoundException fnfe){
    // process errors
  }catch(IOException ioe){
    // process errors
  }
  return lines;
}

getResource() was working fine with the resources files placed in src/main/resources only. To get a file which is at the path other than src/main/resources say src/test/java you need to create it exlicitly.

the following example may help you

import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.net.URISyntaxException;
import java.net.URL;

public class Main {
	public static void main(String[] args) throws URISyntaxException, IOException {
		URL location = Main.class.getProtectionDomain().getCodeSource().getLocation();
		BufferedReader br = new BufferedReader(new FileReader(location.getPath().toString().replace("/target/classes/", "/src/test/java/youfilename.txt")));
	}
}

You can use the com.google.common.io.Resources.getResource to read the url of file and then get the file content using java.nio.file.Files to read the content of file.

URL urlPath = Resources.getResource("src/main/resource");
List<String> multilineContent= Files.readAllLines(Paths.get(urlPath.toURI()));

if you are loading file in static method then ClassLoader classLoader = getClass().getClassLoader(); this might give you an error.

You can try this e.g. file you want to load from resources is resources >> Images >> Test.gif

import org.springframework.core.io.ClassPathResource;
import org.springframework.core.io.Resource;

Resource resource = new ClassPathResource("Images/Test.gif");

	File file = resource.getFile();

To read the files from src/resources folder then try this :

DataSource fds = new FileDataSource(getFileHandle("images/sample.jpeg"));

public static File getFileHandle(String fileName){
       return new File(YourClassName.class.getClassLoader().getResource(fileName).getFile());
}

in case of non static reference:

return new File(getClass().getClassLoader().getResource(fileName).getFile());

Does the code work when not running the Maven-build jar, for example when running from your IDE? If so, make sure the file is actually included in the jar. The resources folder should be included in the pom file, in <build><resources>.

The following class can be used to load a resource from the classpath and also receive a fitting error message in case there's a problem with the given filePath.

import java.io.InputStream;
import java.nio.file.NoSuchFileException;

public class ResourceLoader
{
	private String filePath;

	public ResourceLoader(String filePath)
	{
		this.filePath = filePath;

		if(filePath.startsWith("/"))
		{
			throw new IllegalArgumentException("Relative paths may not have a leading slash!");
		}
	}

	public InputStream getResource() throws NoSuchFileException
	{
		ClassLoader classLoader = this.getClass().getClassLoader();

		InputStream inputStream = classLoader.getResourceAsStream(filePath);

		if(inputStream == null)
		{
			throw new NoSuchFileException("Resource file not found. Note that the current directory is the source folder!");
		}

		return inputStream;
	}
}

I got it work on both running jar and in IDE by writing as

InputStream schemaStream = 
      ProductUtil.class.getClassLoader().getResourceAsStream(jsonSchemaPath);
byte[] buffer = new byte[schemaStream.available()];
schemaStream.read(buffer);

File tempFile = File.createTempFile("com/package/schema/testSchema", "json");
tempFile.deleteOnExit();
FileOutputStream out = new FileOutputStream(tempFile);
out.write(buffer);

this.getClass().getClassLoader().getResource("filename").getPath()

My file in the test folder could not be found even though I followed the answers. It got resolved by rebuilding the project. It seems IntelliJ did not recognize the new file automatically. Pretty nasty to find out.

I get it to work without any reference to "class" or "ClassLoader".

Let's say we have three scenarios with the location of the file 'example.file' and your working directory (where your app executes) is home/mydocuments/program/projects/myapp:

a)A sub folder descendant to the working directory: myapp/res/files/example.file

b)A sub folder not descendant to the working directory: projects/files/example.file

b2)Another sub folder not descendant to the working directory: program/files/example.file

c)A root folder: home/mydocuments/files/example.file (Linux; in Windows replace home/ with C:)

1. Get the right path: a)String path = "res/files/example.file"; b)String path = "../projects/files/example.file" b2)String path = "../../program/files/example.file" c)String path = "/home/mydocuments/files/example.file"

Basically, if it is a root folder, start the path name with a leading slash. If it is a sub folder, no slash must be before the path name. If the sub folder is not descendant to the working directory you have to cd to it using "../". This tells the system to go up one folder.

2. Create a File object by passing the right path:

   File file = new File(path);

3. You are now good to go:

   BufferedReader br = new BufferedReader(new FileReader(file));