How to get the path of src/test/resources directory in JUnit?

Question

I know I can load a file from src/test/resources with:

getClass().getResource("somefile").getFile()

But how can I get the full path to the src/test/resources directory, i.e. I don't want to load a file, I just want to know the path of the directory?

Answer 1

You don't need to mess with class loaders. In fact it's a bad habit to get into because class loader resources are not java.io.File objects when they are in a jar archive.

Maven automatically sets the current working directory before running tests, so you can just use:

    File resourcesDirectory = new File("src/test/resources");

resourcesDirectory.getAbsolutePath() will return the correct value if that is what you really need.

I recommend creating a src/test/data directory if you want your tests to access data via the file system. This makes it clear what you're doing.

Answer 2

Try working with the ClassLoader class:

ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("somefile").getFile());
System.out.println(file.getAbsolutePath());

A ClassLoader is responsible for loading in classes. Every class has a reference to a ClassLoader. This code returns a File from the resource directory. Calling getAbsolutePath() on it returns its absolute Path.

Javadoc for ClassLoader: http://docs.oracle.com/javase/7/docs/api/java/lang/ClassLoader.html

Answer 3

I would simply use Path from Java 7

Path resourceDirectory = Paths.get("src","test","resources");

Neat and clean!

Answer 4

If it's a spring project, we can use the below code to get files from src/test/resource folder.

File file = ResourceUtils.getFile(this.getClass().getResource("/some_file.txt"));

Answer 5

I have a Maven3 project using JUnit 4.12 and Java8. In order to get the path of a file called myxml.xml under src/test/resources, I do this from within the test case:

@Test
public void testApp()
{
    File inputXmlFile = new File(this.getClass().getResource("/myxml.xml").getFile());
    System.out.println(inputXmlFile.getAbsolutePath());
    ...
}

Tested on Ubuntu 14.04 with IntelliJ IDE. Reference here.

Note

The prepending / symbol is necessary, because Class.getResource(String) won't necessarily reveal entire file path (missing) along with FileNotFoundException.

Answer 6

All content in src/test/resources is copied into target/test-classes folder. So to get file from test resources during maven build you have to load it from test-classes folder, like that:

Paths.get(
    getClass().getProtectionDomain().getCodeSource().getLocation().toURI()
).resolve(
    Paths.get("somefile")
).toFile()

Break down:

1. getClass().getProtectionDomain().getCodeSource().getLocation().toURI() - give you URI to target/test-classes.
2. resolve(Paths.get("somefile")) - resolves someFile to target/test-classes folder.

Original anwser is taken from this

Answer 7

There are differences and constraints in options offered by @Steve C and @ashosborne1. They must be specified, I believe.

When can we can use: File resourcesDirectory = new File("src/test/resources");?

• 1 When tests are going to be run via maven only but not via IDE.
• 2.1 When tests are going to be run via maven or
• 2.2 via IDE and only one project is imported into IDE. (I use “imported” term, cause it is used in IntelliJ IDEA. I think users of eclipse also import their maven project). This will work, cause working directory when you run tests via IDE is the same as your project.
• 3.1 When tests are going to be run via maven or
• 3.2 via IDE, and more than one projects are imported into IDE (when you are not a student, you usually import several projects), AND before you run tests via IDE, you manually configure working directory for your tests. That working directory should refer to your imported project that contains the tests. By default, working directory of all projects imported into IDE is only one. Probably it is a restriction of IntelliJ IDEA only, but I think all IDEs work like this. And this configuration that must be done manually, is not good at all. Working with several tests existing in different maven projects, but imported into one big “IDE” project, force us to remember this and don’t allow to relax and get pleasure from your work.

Solution offered by @ashosborne1 (personally I prefer this one) requires 2 additional requirements that must be done before you run tests. Here is a list of steps to use this solution:

• Create a test folder (“teva”) and file (“readme”) inside of “src/test/resources/”:

  src/test/resources/teva/readme

  File must be created in the test folder, otherwise, it will not work. Maven ignores empty folders.

• At least once build project via mvn clean install. It will run tests also. It may be enough to run only your test class/method via maven without building a whole project. As a result your test resources will be copied into test-classes, here is a path: target/test-classes/teva/readme

• After that, you can access the folder using code, already offered by @ashosborne1 (I'm sorry, that I could not edit this code inside of this list of items correctly):

public static final String TEVA_FOLDER = "teva"; ... 
URL tevaUrl = YourTest.class.getClassLoader().getResource(TEVA_FOLDER); 
String tevaTestFolder = new File(tevaUrl.toURI()).getAbsolutePath();

Now you can run your test via IDE as many times as you want. Until you run mvn clean. It will drop the target folder.

Creating file inside a test folder and running maven first time, before you run tests via IDE are needed steps. Without these steps, if you just in your IDE create test resources, then write test and run it via IDE only, you'll get an error. Running tests via mvn copies test resources into target/test-classes/teva/readme and they become accessible for a classloader.

You may ask, why do I need import more than one maven project in IDE and why so many complicated things? For me, one of the main motivation: keeping IDA-related files far from code. I first create a new project in my IDE. It is a fake project, that is just a holder of IDE-related files. Then, I import already existing maven projects. I force these imported projects to keep IDEA files in my original fake project only. As a result I don't see IDE-related files among the code. SVN should not see them (don't offer to configure svn/git to ignore such files, please). Also it is just very convenient.

Answer 8

The simplest and clean solution I uses, suppose the name of the test class is TestQuery1 and there is a resources directory in your test folder as follows:

├── java
│   └── TestQuery1.java
└── resources
    └── TestQuery1
        ├── query.json
        └── query.rq

To get the URI of TestQuery1 do:

URL currentTestResourceFolder = getClass().getResource("/"+getClass().getSimpleName());

To get the URI of one of the file TestQuery1, do:

File exampleDir = new File(currentTestResourceFolder.toURI());
URI queryJSONFileURI = exampleDir.toURI().resolve("query.json");

Answer 9

You can get where you are through

new File(".").getAbsolutePath()

then you can derive path to src/test/resources
usually it is just

new File("src/test/resources")

Answer 10

You can't use a file from a resource folder for tests in a common case. The reason is that resource files in the resource folder are stored inside a jar. So they don't have a real path in the file system.

The most simple solution can be:

1. Copy a file from resources to the temporary folder and get a path to that temporary file.
2. Do tests using a temporary path.
3. Delete the temporary file.

TemporaryFolder from JUnit can be used to create temporary files and delete it after test is complited. Classes from guava library are used to copy a file form resource folder.

Please, notice that if we use a subfolder in the resources folder, like good one, we don't have to add leading / to the resource path.

public class SomeTest {

    @Rule
    public TemporaryFolder tmpFolder = new TemporaryFolder();


    @Test
    public void doSomethinge() throws IOException {
        File file = createTmpFileFromResource(tmpFolder, "file.txt");
        File goodFile = createTmpFileFromResource(tmpFolder, "good/file.txt");

        // do testing here
    }

    private static File createTmpFileFromResource(TemporaryFolder folder,
                                                  String classLoaderResource) throws IOException {
        URL resource = Resources.getResource(classLoaderResource);

        File tmpFile = folder.newFile();
        Resources.asByteSource(resource).copyTo(Files.asByteSink(tmpFile));
        return tmpFile;
    }

}

Answer 11

With Spring you could easily read it from the resources folder (either main/resources or test/resources):

For example create a file: test/resources/subfolder/sample.json

@Test
public void testReadFile() {
    String json = this.readFile("classpath:subfolder/sample.json");
    System.out.println(json);
}

public String readFile(String path) {
    try {
        File file = ResourceUtils.getFile(path);
        return new String(Files.readAllBytes(file.toPath()));
    } catch (IOException e) {
        e.printStackTrace();
    }

    return null;
}

Answer 12

Wow the correct answer is not here yet!

MyClass.class.getResource("/somefile");
MyClass.class.getResourceAsStream("/somefile");

https://javachannel.org/posts/how-to-access-static-resources/

Answer 13

Use the following to inject Hibernate with Spring in your unit tests:

@Bean
public LocalSessionFactoryBean getLocalSessionFactoryBean() {
    LocalSessionFactoryBean localSessionFactoryBean = new LocalSessionFactoryBean();
    localSessionFactoryBean.setConfigLocation(new ClassPathResource("hibernate.cfg.xml"));
    localSessionFactoryBean.setPackagesToScan("com.example.yourpackage.model");
    return localSessionFactoryBean;
}

If you don't have the hibernate.cfg.xml present in your src/test/resources folder it will automatically fall back to the one in your src/main/resources folder.

Answer 14

Use .getAbsolutePath() on your File object.

getClass().getResource("somefile").getFile().getAbsolutePath()

Answer 15

List<String> lines = Files.readAllLines(Paths.get("src/test/resources/foo.txt"));
lines.forEach(System.out::println);

Answer 16

With Spring, you can use this:

import org.springframework.core.io.ClassPathResource;

// Don't worry when use a not existed directory or a empty directory
// It can be used in @before
String dir = new ClassPathResource(".").getFile().getAbsolutePath()+"/"+"Your Path";

Answer 17

Here is some helper functions when using Kotlin language:

private val resourceAccessor = object {}

fun getResourceAsPath(name: String): Path = resourceAccessor
    .javaClass
    .getResource("/$name")!!
    .toURI()
    .toPath()

fun getResourceAsURL(name: String): URL {
    val testResourcesDirectory = Path("src/test/resources").absolutePathString()
    return URL("file:$testResourcesDirectory/$name")
}

class MyTests {
    @Test
    fun myTest() {
        val myResource = getResourceAsPath("file.txt")
        val absolutePath = myResource.absolutePathString()
    }
}