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

Question

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

possible duplicate of [Java - class.getResource returns null](http://stackoverflow.com/questions/4301329/java-class-getresource-returns-null) — maja, Apr 11 '15 at 18:23

score 29 · Answer 1 · answered Oct 16 '16 at 00:28

29

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~

answered Oct 16 '16 at 00:28

MewX

4,232
1
29
38

1

did not work for me, i changed the Compiler resource patterns. Made the resource folder both as the root folder of my project and then the working folder where my Class is and still didn't work! please help. I am using intellij – Metin Dagcilar Dec 20 '17 at 00:11
@AVERAGE Hi, I don't think that would be a problem. Have you tried re-creating a project, and redo your operation? If it still doesn't work, please post your detailed operation steps. That would be helpful for solving your issue. – MewX Dec 28 '17 at 10:25
for me, it worked if it's "src/resources" next to "src/main" path, but got null if in "resources" path – BingLi224 Jan 25 '20 at 04:24
My sample program start working and that sample "HELLO" window is appearing but even though now the error is still there i.e->"Argument 'getClass().getResource("sample.fxml")' might be null" – Pranshu_Taneja Jun 27 '21 at 06:40

score 25 · Answer 2 · edited Apr 11 '15 at 18:23

25

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.

edited Apr 11 '15 at 18:23

maja

17,250
17
82
125

answered Oct 12 '14 at 17:51

stacky

800
6
18

2

I add fxml files using this pattern ?*.fxml but the result is the same! – BilalDja Oct 12 '14 at 18:13
Why "?*.fxml" and not "\*.fxml" or "view:*" - all files and folders within the directory view. see : https://www.jetbrains.com/idea/webhelp/compiler.html#pattern – stacky Oct 12 '14 at 18:21
I follow the same pattern exist in the Resources field – BilalDja Oct 12 '14 at 20:27
4

it doesn't make any change! – BilalDja Oct 12 '14 at 20:31
I had same kind of issue with Jidea 14.1.1 and after updating resource pattern it solve the problem. :) – Coder Aug 24 '15 at 08:49
If adding the pattern did not help, try the answer in http://stackoverflow.com/a/4301353/954439 – srgsanky Feb 08 '17 at 21:21
2

@BilalDja did you manage to fix this problem? As i am experiencing the same issue right now – Metin Dagcilar Dec 20 '17 at 00:09
1

Same here: I'm having this same problem I'm not managing to get it sorted out. I've already tried to change the Resource Patterns. – Davide3i Dec 22 '17 at 09:51

score 25 · Answer 3 · edited Jun 20 '20 at 09:12

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.

`one slash before their names` thanks, I was missing the leading slash... — mihca, Jul 21 '22 at 12:12

score 4 · Answer 4 · answered Jan 20 '18 at 08:48

4

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

answered Jan 20 '18 at 08:48

Max Yan

41
3

1

Other than the methods mentioned above, this is a key point if use Maven. Never forget to load the pom.xml file and write the resource node in build node. – Sinri Edogawa Jul 24 '18 at 08:05
applies to Kotlin as well. – Raj Apr 16 '19 at 02:05
It worked for me. Look here: https://maven.apache.org/plugins/maven-resources-plugin/examples/resource-directory.html – chirag soni Dec 01 '22 at 13:38

score 4 · Answer 5 · answered Dec 06 '20 at 22:53

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>()

or

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

score 2 · Answer 6 · answered Jun 18 '21 at 08:49

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.

Another pitfall of IntelliJ, if you try to create directory structure as explained above, is that IntelliJ will automatically convert them to package structure (my.tests). This happens in IntelliJ's default view - Project view. The only way to create directories under resources is to switch to the Project Files view. Only then you are able to crate a proper directory structure... — caytekin, Sep 15 '22 at 08:48

kinORnirvana · Answer 7 · 2020-05-26T08:27:36.823

1

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.

edited May 26 '20 at 08:27

answered May 14 '20 at 16:03

kinORnirvana

1,667
2
17
22

Сергей · Answer 8 · 2022-11-07T06:32:20.893

1

In my case, ClassLoader helped

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

edited Nov 07 '22 at 06:32

answered Feb 24 '22 at 15:42

Сергей

111
4

Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Feb 25 '22 at 08:54
Furthermore, there must not be a `/` at the beginning of `fileName`. – koppor Nov 06 '22 at 15:34

score 0 · Answer 9 · 2020-02-09T18:12:56.297

0

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---->

Fixed similar issue today by adding resources folder into Resource Tab in IntelliJ IDE

Hope this helps. Also, details tutorial.

edited Feb 09 '20 at 18:12

answered Feb 08 '20 at 17:40

1

Thanks for answering, I don't proved if it's the correct answer ;) Linking images is not a well seen behaviour on stackoverflow. Can you describe what you have done? Perhaps in a list style: 1.) click here 2.) press the button or arrow style "ressources->edit" – nologin Feb 08 '20 at 17:54
1

Thanks for suggestion @nologin - I'm new to stackoverflwo community.. Updated answer. – Feb 09 '20 at 18:13

score 0 · Answer 10 · edited May 03 '21 at 09:27

0

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.

edited May 03 '21 at 09:27

Tomerikoo

18,379
16
47
61

answered May 03 '21 at 08:50

Dion

78
1
9

score 0 · Answer 11 · answered Sep 24 '21 at 11:47

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 .

score 0 · Answer 12 · answered Mar 08 '22 at 21:02

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/

score -1 · Answer 13 · answered Dec 20 '17 at 13:38

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

to

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")

score -1 · Answer 14 · edited Dec 28 '21 at 05:55

-1

Use this syntax:

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

edited Dec 28 '21 at 05:55

U13-Forward

69,221
14
89
114

answered Dec 28 '21 at 05:52

هذا ديننا الإسلام

1

completely wrong - periods are not supported in resource lookup! – kleopatra Feb 25 '22 at 07:54

score -2 · Answer 15 · answered Dec 20 '17 at 13:10

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());

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

15 Answers15

TL; DR;

Long Story;

Linked