Reputation: 1092

IntelliJ IDEA - getClass().getResource("...") return null

I'm using IntelliJ IDEA 13.1.5, I used to work with Eclipse. I'm working on JavaFX application, I try to load FXML file within my MainApp class using getClass().getResource(). I read the documentation and I try several idea, at the end I have null.

This is the hierarchy :

dz.bilaldjago.homekode.MainApp.java

dz.bilaldjago.homekode.view.RootLayout.FXML

This is the code snippet I used:

FXMLLoader loader = new FXMLLoader();
loader.setLocation(getClass().getResource("view/RootLayout.fxml"));

I tried other solution such giving the url from the root and using the classLoader

the result is the same. Any idea please

Upvotes: 46

Answers (17)

Jason Crosby

Reputation: 3583

My problem was that I added it to the wrong directory src. It needs to be in src/main

Upvotes: 0

user10713884

Reputation:

As per suggestion, updated answer.

Step-1

Right click on project
Click on Mark Directory as
Click on Sources Root

Step-2

Click on File in Menu Bar
Click on Project Structure… to open settings panel

Step-3

Click on Modules tab
As you see there isn’t any resources folder added as Content Root
We need to add resources folder into it

Step-4

Make sure to click on resource folder
Click on Resources tab after that
You will see resources folder added as Resource Folders in right panel

Re-run your java program and now it should work.

<---Previous Answer---->

Upvotes: 0

Jérôme B

Reputation: 427

I solved this issue using method getRessource from object ClassLoaderUtils in Apache santuario-java (that was already in my dependency).

https://github.com/apache/santuario-java/blob/trunk/src/main/java/org/apache/xml/security/utils/ClassLoaderUtils.java

Upvotes: 0

Сергей

Reputation: 130

In my case, ClassLoader helped

Class testClass = getClass();
URL url = testClass.getClassLoader().getResource(/fileName);

Upvotes: 1

theplushieisntenough

Reputation: 1

I had this same problem and tried all of the answers to this post. I finally found an answer on a different forum that works. This was the code I put into the build section of pom.xml

<resources>
        <resource>
            <directory>src/main/resources</directory>
            <filtering>true</filtering>
        </resource>
    </resources>

It is important to make sure the directory section is the directory of the resource class, not the main class.

I also changed my compiler settings as was suggested in another comment. I found this answer on a post here: https://javawithus.com/en/classloadergetresource-returns-null/

Upvotes: 0

هذا ديننا الإسلام

Reputation: 1

Use this syntax:

("../view/RootLayout.fxml"))

Upvotes: -1

Mukesh Kumar

Reputation: 1

I used below code to access the file from resource folder in spring application. I used classLoader.getResourceAsStream(FileName) instead of getClass().getResourceAsStream(FileName), Hope this would help to access the file residing in resource folder.

String fileName = "./file.txt";
ClassLoader classLoader = getClass().getClassLoader();
InputStream inputStream = classLoader.getResourceAsStream(fileName);

I resolved file access so thought to share with all .

Upvotes: 0

Jaco Van Niekerk

Reputation: 4192

Very frustrating. What also happens is that the directory containing the resources are sometimes created as:

 └── resources
     └── my.tests             
           ├── t1.txt
           └── t2.txt

instead of:

 └── resources
     └── my
          └── tests
              ├── t1.txt
              └── t2.txt

The solution is to just manually recreate the directory structure so that it resembles the second example.

I also recommend to to troubleshoot your tests on the command line (with mvn clean test) if you're using maven to eliminate the chances of this being an IDE issue.

Upvotes: 3

Dion

Reputation: 100

In Project Structure > Project you have to make sure Project compiler output: has its value filled in. In my case it did not. Point it to the ./target or ./bin (whatever you have) directory in your project.

Upvotes: 0

Rasoul Miri

Reputation: 12222

first, you need to set src and resource folders for IntelliJ.

see the icon near the java(blue) and resources(yellow 4 lines) folder.

right-click on java folder and select sources root

and right-click on java folder and select resource root

after that create the same package in the java folder and resource folder.

for example: packages -> org.example

├─ src
   ├─ main
       ├─ java
       |  └─ com
       |      └─ example
       |           └─ A.class
       |
       └─ resources  
          └─ com
              └─ example
                    └─ fxml
                         └─ AFXML.fxml

in A.class you can use this.

Java

 FXMLLoader loader = new FXMLLoader(getClass().getResource("fxml/AFXML.fxml"));
 Parent root = loader.load();

Kotlin

 val loader = FXMLLoader(javaClass.getResource("fxml/AFXML.fxml"))
 val root = loader.load<Parent>()

 val loader = FXMLLoader(A::class.java.getResource("fxml/AFXML.fxml"))
 val root = loader.load<Parent>()

Upvotes: 4

kinORnirvana

Reputation: 1715

If your project is Gradle, check Settings -> Build, Execution, Deployment -> Gradle -> Build and run section. Ensure that "Build and run using" option is "Gradle".

Intellij IDEA's compiler doesn't copy "resources" to build dir nor to tests classpath. Also, it unexpectedly uses dummy "out" dir instead of "build".

UPDATED: only 1 option "Build and run using" is enough to set to Gradle.

Upvotes: 2

AlwaysLearner

Reputation: 473

TL; DR;

Put your resources in resources folder.
Use them with one slash before their names: getClass().getResource("/myfont.ttf");

Long Story;

If you are using Intellij IDEA and you created a Maven project, you should put your resources in resources folder (as marked as resource root by intellij itself) and these resource go to the root of your compiled app.

I mean, /resources/myfont.ttf will go to /myfont.ttf in the resulting build.

So you should get it via /myfont.ttf and not myfont.ttf. Use it like this:

getClass().getResource("/myfont.ttf");

No need to change anything else. Just this one helped me.

Upvotes: 27

Max Yan

Reputation: 41

if your project is a maven project, check the target code to see whether your .fxml file exist there. if it's not there ,just add

<resources>
        <resource>
            <directory>src/main/java</directory>
            <includes>
                <include>**/*.xml</include>
            </includes>
            <filtering>true</filtering>
        </resource>
    </resources>

in your pom.xml

Upvotes: 4

Joop Eggen

Reputation: 109613

Windows is case-sensitive, the rest of the world not. Also an executable java jar (zip format) the resource names are case sensitive.

Best rename the file

view/RootLayout.FXML

view/RootLayout.fxml

This must be done by moving the original file away, and creating a new one.

Also compile to a jar, and check that the fxml file was added to the jar (zip file). When not IntelliJ resource paths are treated by an other answer.

By the way this is path relative to the package path of getClass(). Be aware if you extended this class the full path changes, better use:

MainApp.class.getResource("view/RootLayout.fxml")

Upvotes: -1

Metin Dagcilar

Reputation: 554

I gave up trying to use getClass().getResource("BookingForm.css"));

Instead I create a File object, create a URL from that object and then pass it into getStyleSheets() or setLocation()File file = new File(System.getProperty("user.dir").toString() + "/src/main/resources/BookingForm.css");

scene.getStylesheets().add(folder.toURI().toURL().toExternalForm());

Upvotes: -2

MewX

Reputation: 4312

I solved this problem by pointing out the resource root on IDEA.

Right click on a directory (or just the project name) -> Mark directory As -> Resource Root.

Recompile & rejoice :P Hope this working for you~

Upvotes: 29

stacky

Reputation: 820

For those who use Intellij Idea: check for Settings -> Compiler -> Resource patterns.

The setting contains all extensions that should be interpreted as resources. If an extension does not comply to any pattern here, class.getResource will return null for resources using this extension.

Upvotes: 26

IntelliJ IDEA - getClass().getResource(&quot;...&quot;) return null

Answers (17)

TL; DR;

Long Story;

Related Questions

IntelliJ IDEA - getClass().getResource("...") return null