How do I open the JavaFX FileChooser from a controller class?

旧时模样 提交于 2019-11-27 08:15:22

For any node in your scene (for example, the root node; but any node you have injected with @FXML will do), do

chooser.showOpenDialog(node.getScene().getWindow());

You don't have to stick with the Stage created in the Application you can either:

@FXML protected void locateFile(ActionEvent event) {
    FileChooser chooser = new FileChooser();
    chooser.setTitle("Open File");
    File file = chooser.showOpenDialog(new Stage());
}

Or if you want to keep using the same stage then you have to pass the stage to the controller before:

    FXMLLoader loader = new FXMLLoader(getClass().getResource("yourFXMLDocument.fxml"));
    Parent root = (Parent)loader.load();
    MyController myController = loader.getController();
    myController.setStage(stage);

and you will have the main stage of the Application there to be used as you please.

From a menu item

public class SerialDecoderController implements Initializable {

  @FXML
  private MenuItem fileOpen;

  @Override
  public void initialize(URL url, ResourceBundle rb) {
    // TODO
 }    


public void fileOpen (ActionEvent event) {

    FileChooser fileChooser = new FileChooser();
    fileChooser.setTitle("Open Resource File"); 
    fileChooser.showOpenDialog(fileOpen.getParentPopup().getScene().getWindow());

}

Alternatively, what worked for me: simply put null.

@FXML
private void onClick(ActionEvent event) {
    File file = fileChooser.showOpenDialog(null);
    if (file != null) {
       //TODO
    }
}
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