HTTP請求後在單獨的線程上更改呈現屏幕

我的目標是在遊戲無法到達後端時更改用戶的屏幕。我的代碼按預期執行，除非屏幕永遠不會改變。這裏的初始呼叫：HTTP請求後在單獨的線程上更改呈現屏幕

timer.testTimeToServer(api, game);

這裏的計時器對象的類。我把（我的網址）代替我的後端的實際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") 
      } 
     } 
    }); 
    } 
}

這裏是在API對象的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); 
}

我很爲難，因爲代碼打印出「應該切換屏幕「，所以它的行爲像預期的那樣，除了遊戲被凍結並且屏幕切換從未實際發生。

來源

2016-02-10 Spencer Sutton

OpenGL不能很好地處理多線程。 –

您是否嘗試在更改屏幕的渲染方法內創建驗證？在java中，當你將變量傳遞給方法時，它們是實例的副本，所以在這個方法中調用game.setScreen 對原始實例不起作用，這不是像C，C++這樣的指針。您需要以靜態方式訪問遊戲對象，或者在渲染方法中驗證它並更改屏幕。哦，你需要調用dispose（）;更換屏幕之後... – Hllink

懶惰的方法：

在遊戲主類：

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") 
      } 
     } 
    }); 
    } 
}

正確的方式

您可以創建一個呼叫回到ProofOfConcept類的內部，渲染方法的每一幀都會檢查結果並更改屏幕。

來源

2016-02-11 12:51:11 Hllink

不錯，就是這樣！謝謝一堆！ –

很高興幫助^ - ^ – Hllink

HTTP請求後在單獨的線程上更改呈現屏幕

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