JAVA FX-Service.start()不会在单独的线程中运行 [英] JAVA FX - Service.start() wont run in a separate thread

问题描述

我有应用程序，我在primaryStage中开始了自己的场景.

I have and application and i start my scene in the primaryStage.

当用户单击确定"按钮时，我创建一个服务，然后启动它.该服务调用WebService并得到响应. 我已经看到，在调用start()之后，主流不再继续，并且createTask()立即被调用，并且我所有的UI都被阻止了.这是我的代码:

When the user click OK button, I create a Service and after that, I start it. This service make a call to WebService and get me the response. I've seen that after the start() is called, the main flow do not continue and createTask() is called immediately and all my UI is blocked. This is my code:

当我按下按钮时输入代码

@FXML
    private void handleRegisterButton() {

        User newUser = new User();
        newUser.setUsername(usernameTextField.getText());
        newUser.setGroup(groupChoicheBox.getSelectionModel().getSelectedItem());
        newUser.setMachineName(machineNameTextField.getText());

        RegisterUserInvocation service = new RegisterUserInvocation(newUser);

        service.stateProperty().addListener(new InvalidationListener() {
            @Override
            public void invalidated(Observable observable) {
                System.out.println("Task value " + service.getState());
                if(service.getState().equals(Worker.State.SUCCEEDED))
                    System.out.println("SUCCEEDED");
            }
        });

        service.start();

        this.mainApp.getPrimaryStage().close();
    }

RegisterUserInvocation代码

public class RegisterUserInvocation extends Service<Integer>{
    private User user;

    public RegisterUserInvocation(User user) {
        this.user = user;
    }

    @Override
    protected Task<Integer> createTask() {
        String url = "http://localhost:8080/LANServer/services/Rest/user/";
        Client client = ClientBuilder.newClient();      
        Integer resp = client.target(url).request().post(Entity.entity(user, MediaType.APPLICATION_JSON), Integer.class);

        return new Task<Integer>() {
            @Override
            protected Integer call() throws Exception {
                return resp;
            }
        };
    }
}

推荐答案

我发现我错了，这是 RegisterUserInvocation 的新代码:

I've found where i wrong, this is the new code of RegisterUserInvocation:

public class RegisterUserInvocation extends Service<Integer>{
    private User user;

    public RegisterUserInvocation(User user) {
        this.user = user;
    }

    @Override
    protected Task<Integer> createTask() {
        return new Task<Integer>() {
            @Override
            protected Integer call() throws Exception {
                String url = "http://localhost:8080/LANServer/services/Rest/user/";
                Client client = ClientBuilder.newClient();      
                Integer resp = client.target(url).request().post(Entity.entity(user, MediaType.APPLICATION_JSON), Integer.class);

                return resp;
            }
        };
    }
}

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