How do I load a file from resource folder?

Question

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.

Eldwen

Doesn't work how? What is your error? – Deflagrate

try this this.getClass().getResource("/test.csv") – Ehlers

#3891875 – Ordinary

@Ehlers it worked (cause this will give absolute path url in return ) however the moment i make jar file it's not working as its inside a jar and absolute path becomes invalid, is there a way to play with relative path itself – Dougall

@SRy, somewhere between now and 2013, this seems to have been fixed. I am today able to load root resources without the initial /. However, I do getClass().getClassLoader().getResourceAsStream(filename)... maybe that's the difference? – Nunley

Answer 1

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.

Salverform

thx for the answer, could you please explain why we should use this very specific loader but not that of the class ? – Sofa

great, I'm so stupid that I was using Object.class.getClassLoader();, from a static context which didn't work - this suggestion does - well almost, it injects %20 for spaces which gives me a FileNotFoundException – Floc

@Floc Because getResource returns a URL, not a file. The getFile method of java.net.URL does not convert a URL to a file; it just returns the path and query portions of the URL. You shouldn't even try to convert it to a File; just call openStream and read from that. – Semiramis

Check this project, solves resources folder scanning: github.com/fraballi/resources-folder-scanner – Barber

to convert to string in one line: new String(this.getClass().getResourceAsStream("/my_file_in_resources.bzip").readAllBytes()) – Chartres

The one liner suggested by @Chartres only works in Java 9 or higher, see #54269711 – Solis

Answer 2

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

Spangler

For those seeking sanity, this post brings you all the way to getting the resource as a String https://mcmap.net/q/36074/-how-do-i-read-convert-an-inputstream-into-a-string-in-java – Marxism

The "/" is the key. – Melodramatic

Answer 3

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);

Skippet

Should the InputStream not be closed? – Emanate

This answer works on a war file not for a jar file. – Not

Answer 4

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)

Traumatize

The docs say that this about the getResource method: 'getResource(java.lang.String)' is declared in unstable class 'com.google.common.io.Resources' marked with @Beta. Should people really be using methods from an unstable class? – Oppressive

what is this in your usage example? why people cannot attach the whole code needed? – Fimble

Answer 5

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()));

Cark

Answer 6

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

Varico

I must use "/test.csv" (note the slash). – Marivaux

Answer 7

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.

Disruption

Thanks, saved me from thinking I was losing my mind! – Nicolais

Answer 8

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.

Frequently

Programmers should always be concerned about performance. While premature optimization is certainly to be avoided, knowing the more efficient approach to take is always good. It's like knowing the difference between LinkedList and ArrayList and when to use one or the other. – Blueing

@Marvo: Programmers must always be concerned about quality, capabilities and ease of maintenance, performance is at the queue. – Albumenize

Answer 9

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!

Rambunctious

Can you tell us the full name of Font class? – Beaufort

Answer 10

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;
}

Halden

Answer 11

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()));

Jenkins

Answer 12

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")));
    }
}

Workday

Answer 13

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

Macaco

Answer 14

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();

Acidhead

Answer 15

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());

Lemar

Answer 16

For Kotlin I am doing the following:

1. I created an assets folder in my project (for Android studio go to File -> New-> Folder -> Assets Folder)
2. Copied my assets (.txt files) to this folder
3. Then I used the code below

Import these items:

    import android.content.Context
    import android.util.Log
    import java.io.FileNotFoundException

I created a function like this:

    fun loadBook(context: Context) {
       val assetManager = context.assets
       try {
          val inputStream = assetManager?.open("19-Psalms.usfm")
          val lineList = mutableListOf<String>()

          inputStream?.bufferedReader()?.forEachLine { lineList.add(it) }
          lineList.forEach{println(">  $it")}
      }
      catch (e:FileNotFoundException) {
          Log.e("FileNotFoundException","File not found")
      }
    }
 

Hola

Answer 17

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>.

Integrand

When using eclipse and running code from IDE itself. How can we load resources located at "/src/test/resources" in Java code specifically in unit tests. Consider a standard maven project structure. – Corruption

Answer 18

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;
    }
}

Hydrodynamics

Answer 19

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);

Subsist

How does your file structure look like? – Cascarilla

Answer 20

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.

Sprinkle

Answer 21

I am currently doing it this way.

This is my structure.

final String path =   "src/test/resources/archivos/B-163.txt";
final Path filePath = Path.of( path );
final String documento = Files.readString( filePath );
         
log.debug("DOCUMENTO: {}" , document); 

I am working with Java 11.

Holiday

Answer 22

For Kotlin:

import java.nio.charset.StandardCharsets
import java.nio.file.Files
import java.nio.file.Paths

val lines = Files.readAllLines(
    Paths.get((object {}).javaClass.getResource("videos.txt").toURI()),
    StandardCharsets.UTF_8
)

lines // will contain list of lines in a file

Impeller

Answer 23

This worked pretty fine for me :

InputStream in = getClass().getResourceAsStream("/main/resources/xxx.xxx");
InputStreamReader streamReader = new InputStreamReader(in, StandardCharsets.UTF_8);
BufferedReader reader = new BufferedReader(streamReader);
String content = "";
for (String line; (line = reader.readLine()) != null;) {
    content += line;
}

Mild

Answer 24

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));

Melanism