How to reference javafx fxml files in resource folder?

Generic Resource Lookup Information

This answer discusses FXML location lookup, which is really just a subset of the generic resource lookup task in Java. The resource location is passed by the calling program as input to the FXMLLoader, so the resource lookup itself is part of your calling application code and not the FXMLLoader.

For thorough coverage of generic resource information (including recommended troubleshooting steps) for Java/JavaFX applications, please refer to:

• How do I determine the correct path for FXML files, CSS files, Images, and other resources needed by my JavaFX Application?

Example usage

FXMLLoader loader = new FXMLLoader();loader.setLocation(getClass().getResource("/main.fxml"));Parent content = loader.load(); 

Location resolution options

1. Put all of your fxml in the src/main/resources directory.

   loader.setLocation(getClass().getResource("/main.fxml"));

2. Put all of your fxml in a src/main/resources/fxml directory.

   loader.setLocation(getClass().getResource("/fxml/main.fxml"));

3. Place fxml in a corresponding resource directory; e.g. src/main/resources/com/mycompany/myapp.

   loader.setLocation(getClass().getResource("main.fxml")); 

The last option assumes that the class from which you are loading the fxml is in the same relative location in the corresponding Java source hierarchy. For example, you might invoke the last load command from source com/mycompany/myapp/Main.java.

FXMLLoader usage recommendations

1. Instantiate an FXMLLoader via new FXMLLoader() rather than usingthe static methods on the FXMLLoader.

   • The static methodsbecome confusing when you want to get values (like instantiatedcontrollers) out of a loader.

2. Set the location on the instantiated FXMLLoader and callload() rather than loading from a stream usingload(stream).

   • Setting a URL based location on the loader allows for resolution ofrelative resources loaded in fxml and css files. Relativeresources do not resolve for a stream based constructor.

3. To derive a location based upon a class, usegetClass().getResource(), as it is URL based, rather thangetClass().getResourceAsStream() which is stream based.

IDE and Build Settings

Ensure that your IDE or build tool is copying the fxml files from the resource directory to the build output directory. For understanding Intellij settings for this, see: How to convert a normal java project in intellij into a JavaFx project.

A note on Java Jigsaw modular applications

See:

• How to access resource using class loader in Java 9

Specifically, don't write:

ComboBoxStyling.class.getClassLoader()    .getResource("/css/styleclass.css");

Instead write:

ComboBoxStyling.class    .getResource("/css/styleclass.css"); 

That is, get the resource from the class directly rather, NOT from the class loader. If you get resources from a class loader rather than the class, then there are additional restrictions on lookup based upon the module configuration. These restrictions can be hard to understand, see the documentation for the Class and ClassLoader getResource methods for information if needed.

You are using FXMLLoader fxmlLoader = new FXMLLoader( getClass().getResource("main.fxml") ); to load your main.fxml. This requires the main.fxml and the java class loading it in same directory/path.If you want to load an fxml file from a path/location different from the java class loading it, you need to use relative path.As @jewelsea mentioned above, you may use the relative path using / character before your main.fxml.So the code that will make it work for you in your case is

FXMLLoader fxmlLoader = new FXMLLoader( getClass().getResource("/main.fxml") );

Open your resources folder in the file explorer of your operation system. Here you will see that you have created a "dominion.application" folder instead of an "application" folder inside of a "dominion" folder. Because of this the "MainController" class is placed in a different package as the fxml file.