Question

我的目标是在游戏无法再到达后端时更改用户的屏幕。我的代码按预期执行，但屏幕永不改变。这是最初的电话：

timer.testTimeToServer(api, game);

这里是计时器对象的类。我把（我的网址）代替我后端的实际IP地址：

public class CustomTimer {
private static final float timeToDrop = 2000;
private float time = 0;
private StopWatch watch = new StopWatch();

public void testTimeToServer(ApiCall api,final proofOfConcept game){
    watch.start();
    api.httpGetWithCallback("(my url)/api/v1/character", new CallBack(){
        @Override
        public void callback(String resp){
            System.out.println("Server Responded");
            time = watch.getTime();
            watch.stop();
            watch.reset();
            if(time > timeToDrop){
                game.setScreen(new GameOverScreen(game, false));
                System.out.println("Should have switched screen")
            }
        }
    });
   }
}

这是api对象中的httpGetWithCallback方法：

public void httpGetWithCallback (final String URL, final CallBack callback){
    Thread th = new Thread(new Runnable(){
                    @Override
                    public void run() {
                                Gdx.app.postRunnable(new Runnable() {

                                         @Override
                                         public void run() {

                                             Net.HttpRequest httpRequest = new Net.HttpRequest(Net.HttpMethods.GET);
                                             httpRequest.setUrl(URL);
                                             httpRequest.setHeader("Content-Type", "application/json");
                                             httpRequest.setTimeOut(timeoutTimeInMilli);
                                             Gdx.net.sendHttpRequest(httpRequest, new Net.HttpResponseListener() {
                                                 @Override
                                                 public void handleHttpResponse(Net.HttpResponse httpResponse) {
                                                     String successValue = httpResponse.getResultAsString();
                                                     if (successValue.contains("\"total_count\": 0"))//wrong credentials
                                                     {
                                                         callback.callback("EMPTY");
                                                     } else//there was a match yo! should probably have a unique conststraint on username. too hard eff it
                                                     {
                                                         callback.callback(successValue);
                                                     }
                                                 }

                                                 @Override
                                                 public void failed(Throwable t) {
                                                     callback.callback("FAILED");
                                                 }

                                                 @Override
                                                 public void cancelled() {
                                                     callback.callback("CANCELLED");
                                                 }
                                             });
                                         }
                                     }
                );
            }
    });

    th.start();
    threads.add(th);
}

我很难过，因为代码打印出来＆＃34;应该切换屏幕＆＃34;所以它表现得像预期的那样，除了游戏被冻结并且屏幕切换实际上从未发生过这样的事实。

Answer 1

懒惰的方式：

在你的主要游戏课上：

public static ProofOfConcept game;

和你的方法

public void testTimeToServer(ApiCall api){
    watch.start();
    api.httpGetWithCallback("(my url)/api/v1/character", new CallBack(){
        @Override
        public void callback(String resp){
            System.out.println("Server Responded");
            time = watch.getTime();
            watch.stop();
            watch.reset();
            if(time > timeToDrop){
                Main.game.setScreen(new GameOverScreen(false));
                System.out.println("Should have switched screen")
            }
        }
    });
   }
}

正确的方式

您可以在ProofOfConcept类中创建一个回调，即render方法的每一帧都会检查结果并更改屏幕。

在HTTP请求后在单独的线程上更改渲染屏幕

1 个答案: