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I've tried numerous methods now, including FilenameUtils.normalize() from commons IO, but I can't seem to be able to get a resource in another folder to get a Java FXML file.
The code is the following
try {
root = FXMLLoader.load(getClass().getResource("../plugin/PluginSelection.fxml"));
} catch (IOException ex) {
Logger.getLogger(QueueOperationsController.class.getName()).log(Level.SEVERE, null, ex);
}
Where the desired FXML file is:
gui
dialogues
plugins
PluginSelection.fxml // desired file
dataset
QueueOperationsController // current class
How do I best get the desired file's URL?
Thank you!
208
In Java, we can use getResourceAsStream or getResource to read a file or multiple files from a resources folder or root of the classpath. The getResourceAsStream method returns an InputStream . // the stream holding the file content InputStream is = getClass(). getClassLoader().
The getResource method finds a resource with the specified name. It returns a URL to the resource or null if it does not find the resource. Calling java.
The simplest approach uses an instance of the java. io. File class to read the /src/test/resources directory by calling the getAbsolutePath() method: String path = "src/test/resources"; File file = new File(path); String absolutePath = file.
You can get resources relative to the Class or the context root. In your example putting / at the start of the string if thats your package structure in your application. Try
getClass().getResource("/gui/dialogues/plugins/PluginSelection.fxml")
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