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.
List<String> lines = Files.readAllLines(Paths.get("src/test/resources/foo.txt"));
lines.forEach(System.out::println);
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");?
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.
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"));
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:
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;
}
}
Use .getAbsolutePath() on your File object.
getClass().getResource("somefile").getFile().getAbsolutePath()
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.
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";
I would simply use Path from Java 7
Path resourceDirectory = Paths.get("src","test","resources");
Neat and clean!
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.
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");
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;
}
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:
getClass().getProtectionDomain().getCodeSource().getLocation().toURI() - give you URI to target/test-classes.resolve(Paths.get("somefile")) - resolves someFile to target/test-classes folder.Original anwser is taken from this
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")
Source: Stackoverflow.com