问同时启动JavaFX应用程序和TomcatServer

EN

public class Start {
    
    public static void main( String args[] )
    {
        try {
            
            Tomcat tomcat = new Tomcat( );
            String webappDirectory = new File( "src/main/webapp" ).getAbsolutePath( );
            tomcat.setPort( 8080 );
            Context context = tomcat.addWebapp( "", webappDirectory );
            Tomcat.addServlet( context, "blockchain", new ServletContainer( new Applications( ) ) );
            context.addServletMappingDecoded( "/blockchain/api/*", "blockchain" );
            tomcat.start( );
            tomcat.getServer( ).await( );
            
            ConfigurationGui config = new ConfigurationGui();
            Stage stage = new Stage();
            config.start(stage);
            
        }
        catch ( Exception e )
        {
            e.printStackTrace( );
        }
    }

}

import javafx.application.Application;
import javafx.scene.Parent;
import javafx.scene.Scene;
import javafx.scene.layout.BorderPane;
import javafx.stage.Stage;

public class ConfigurationGui extends Application {

    @Override
    public void start(Stage primaryStage) throws Exception {
        Parent root = new BorderPane();
        Scene scene = new Scene(root,400,400);
        scene.getStylesheets().add(getClass().getResource("application.css").toExternalForm());
        primaryStage.setScene(scene);
        primaryStage.show();
        
    }

}