I am trying to read a text file which is set in CLASSPATH system variable. Not a user variable.

I am trying to get input stream to the file as below:

Place the directory of file (D:\myDir)in CLASSPATH and try below:

InputStream in = this.getClass().getClassLoader().getResourceAsStream("SomeTextFile.txt");
InputStream in = this.getClass().getClassLoader().getResourceAsStream("/SomeTextFile.txt");
InputStream in = this.getClass().getClassLoader().getResourceAsStream("//SomeTextFile.txt");

Place full path of file (D:\myDir\SomeTextFile.txt)in CLASSPATH and try the same above 3 lines of code.

But unfortunately NONE of them are working and I am always getting null into my InputStream in.

Solution 1

With the directory on the classpath, from a class loaded by the same classloader, you should be able to use either of:

// From ClassLoader, all paths are "absolute" already - there's no context
// from which they could be relative. Therefore you don't need a leading slash.
InputStream in = this.getClass().getClassLoader()
// From Class, the path is relative to the package of the class unless
// you include a leading slash, so if you don't want to use the current
// package, include a slash like this:
InputStream in = this.getClass().getResourceAsStream("/SomeTextFile.txt");

If those aren't working, that suggests something else is wrong.

So for example, take this code:

package dummy;

import java.io.*;

public class Test
    public static void main(String[] args)
        InputStream stream = Test.class.getResourceAsStream("/SomeTextFile.txt");
        System.out.println(stream != null);
        stream = Test.class.getClassLoader().getResourceAsStream("SomeTextFile.txt");
        System.out.println(stream != null);

And this directory structure:


And then (using the Unix path separator as I'm on a Linux box):

java -classpath code:txt dummy.Test



Solution 2

When using the Spring Framework (either as a collection of utilities or container - you do not need to use the latter functionality) you can easily use the Resource abstraction.

Resource resource = new ClassPathResource("com/example/Foo.class");

Through the Resource interface you can access the resource as InputStream, URL, URI or File. Changing the resource type to e.g. a file system resource is a simple matter of changing the instance.

Solution 3

This is how I read all lines of a text file on my classpath, using Java 7 NIO:

import java.nio.charset.Charset;
import java.nio.file.Files;
import java.nio.file.Paths;


    Paths.get(this.getClass().getResource("res.txt").toURI()), Charset.defaultCharset());

NB this is an example of how it can be done. You'll have to make improvements as necessary. This example will only work if the file is actually present on your classpath, otherwise a NullPointerException will be thrown when getResource() returns null and .toURI() is invoked on it.

Also, since Java 7, one convenient way of specifying character sets is to use the constants defined in java.nio.charset.StandardCharsets (these are, according to their javadocs, "guaranteed to be available on every implementation of the Java platform.").

Hence, if you know the encoding of the file to be UTF-8, then specify explicitly the charset StandardCharsets.UTF_8

Solution 4

Please try

InputStream in = this.getClass().getResourceAsStream("/SomeTextFile.txt");

Your tries didn't work because only the class loader for your classes is able to load from the classpath. You used the class loader for the java system itself.

Solution 5

To actually read the contents of the file, I like using Commons IO + Spring Core. Assuming Java 8:

try (InputStream stream = new ClassPathResource("package/resource").getInputStream()) {


InputStream stream = null;
try {
    stream = new ClassPathResource("/log4j.xml").getInputStream();
} finally {

Solution 6

To get the class absolute path try this:

String url = this.getClass().getResource("").getPath();

Solution 7

Somehow the best answer doesn't work for me. I need to use a slightly different code instead.

ClassLoader loader = Thread.currentThread().getContextClassLoader();
InputStream is = loader.getResourceAsStream("SomeTextFile.txt");

I hope this help those who encounters the same issue.

Solution 8

If you use Guava:

import com.google.common.io.Resources;

we can get URL from CLASSPATH:

URL resource = Resources.getResource("test.txt");
String file = resource.getFile();   // get file path 

or InputStream:

InputStream is = Resources.getResource("test.txt").openStream();

Ways to convert an InputStream to a String

Solution 9

To read the contents of a file into a String from the classpath, you can use this:

private String resourceToString(String filePath) throws IOException, URISyntaxException
    try (InputStream inputStream = this.getClass().getClassLoader().getResourceAsStream(filePath))
        return IOUtils.toString(inputStream);

IOUtils is part of Commons IO.

Call it like this:

String fileContents = resourceToString("ImOnTheClasspath.txt");

Solution 10

You say "I am trying to read a text file which is set in CLASSPATH system variable." My guess this is on Windows and you are using this ugly dialog to edit the "System Variables".

Now you run your Java program in the console. And that doesn't work: The console gets a copy of the values of the system variables once when it is started. This means any change in the dialog afterwards doesn't have any effect.

There are these solutions:

  1. Start a new console after every change

  2. Use set CLASSPATH=... in the console to set the copy of the variable in the console and when your code works, paste the last value into the variable dialog.

  3. Put the call to Java into .BAT file and double click it. This will create a new console every time (thus copying the current value of the system variable).

BEWARE: If you also have a User variable CLASSPATH then it will shadow your system variable. That is why it is usually better to put the call to your Java program into a .BAT file and set the classpath in there (using set CLASSPATH=) rather than relying on a global system or user variable.

This also makes sure that you can have more than one Java program working on your computer because they are bound to have different classpaths.

Solution 11

My answer is not exactly what is asked in the question. Rather I am giving a solution exactly how easily we can read a file into out java application from our project class path.

For example suppose a config file name example.xml is located in a path like below:-


and our java executable class file is in the below path:-


now just check in both the above path which is the nearest common directory/folder from where you can access both dev and main directory/folder (com.myproject.server.main - where our applications java executable class is existed) We can see that it is myproject folder/directory which is the nearest common directory/folder from where we can access our example.xml file. Therefore from a java executable class resides in folder/directory main we have to go back two steps like ../../ to access myproject. Now following this, see how we can read the file:-

package com.myproject.server.main;

class Example {

  File xmlFile;

  public Example(){
       String filePath = this.getClass().getResource("../../config/dev/example.xml").getPath();
       this.xmlFile = new File(filePath);

  public File getXMLFile() {
      return this.xmlFile;
   public static void main(String args[]){
      Example ex = new Example();
      File xmlFile = ex.getXMLFile();

Solution 12

If you compile your project in jar file: you can put your file in resources/files/your_file.text or pdf;

and use this code:

import org.slf4j.Logger;
import org.slf4j.LoggerFactory;
import java.io.*;

public class readFileService(){
    private static final Logger LOGGER = LoggerFactory.getLogger(readFileService.class);

    public byte[] getFile(){
        String filePath="/files/your_file";
        InputStream inputStreamFile;
        byte[] bytes;
            inputStreamFile = this.getClass().getResourceAsStream(filePath);
            bytes = new byte[inputStreamFile.available()];
        } catch(NullPointerException | IOException e) {
            LOGGER.error("Erreur read file "+filePath+" error message :" +e.getMessage());
            return null;
        return bytes;

Solution 13

I am using webshpere application server and my Web Module is build on Spring MVC. The Test.properties were located in the resources folder, i tried to load this files using the following:

  1. this.getClass().getClassLoader().getResourceAsStream("Test.properties");
  2. this.getClass().getResourceAsStream("/Test.properties");

None of the above code loaded the file.

But with the help of below code the property file was loaded successfully:


Thanks to the user "user1695166".

Solution 14

Use org.apache.commons.io.FileUtils.readFileToString(new File("src/test/resources/sample-data/fileName.txt"));

Solution 15

Don't use getClassLoader() method and use the "/" before the file name. "/" is very important


Solution 16

import java.io.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;

public class ReadFile

     * * feel free to make any modification I have have been here so I feel you
     * * * @param args * @throws InterruptedException

    public static void main(String[] args) throws InterruptedException {
        // thread pool of 10
        File dir = new File(".");
        // read file from same directory as source //
        if (dir.isDirectory()) {
            File[] files = dir.listFiles();
            for (File file : files) {
                // if you wanna read file name with txt files
                if (file.getName().contains("txt")) {

                // if you want to open text file and read each line then
                if (file.getName().contains("txt")) {
                    try {
                        // FileReader reads text files in the default encoding.
                        FileReader fileReader = new FileReader(
                        // Always wrap FileReader in BufferedReader.
                        BufferedReader bufferedReader = new BufferedReader(
                        String line;
                        // get file details and get info you need.
                        while ((line = bufferedReader.readLine()) != null) {
                            // here you can say...
                            // System.out.println(line.substring(0, 10)); this
                            // prints from 0 to 10 indext
                    } catch (FileNotFoundException ex) {
                        System.out.println("Unable to open file '"
                                + file.getName() + "'");
                    } catch (IOException ex) {
                        System.out.println("Error reading file '"
                                + file.getName() + "'");
                        // Or we could just do this:



Solution 17

you have to put your 'system variable' on the java classpath.