Question

我有一个如下所示的实现，它将动态创建的js返回到webview的shouldInterceptRequest方法，但是我需要在它之前创建一个请求并在返回inputStream之前等待它的答案。我怎么能让它等待呢。

我可以为此准备和同步请求，它解决了问题，但我只是想知道有没有办法在新线程中执行此操作。

MRURLProtocol.mapDomain("gamification.fs", new MRProtocolResponseHandler() {
            @Override
            public InputStream handle(Uri uri) {
             InputStream stream = null;
                myApplication.getChamp().getImpulser().invokeRequest(method, body,new OnWebApiResponseArrived() {
                    @Override
                    public void OnSuccess(Object obj) {
                        String configString = "(function () { if (typeof(window) === 'undefined') ..... %s ....";
                        configString = String.format(configString, obj);
                        stream = new ByteArrayInputStream(configString.getBytes());
                    }

                    @Override
                    public void OnFail(String errMsg) {
                        String configString = "(function () { if (typeof(window) === 'undefined')..... %s ....";
                        configString = String.format(configString, errMsg);
                        stream = new ByteArrayInputStream(configString.getBytes());
                    }
                }); 
                // NEED TO WAIT FOR stream before return

                return stream;
            }
        });

调用请求（以显示它正在处理新线程）

public void invokeRequest(String entryPoint,String body,final OnWebApiResponseArrived callback ){
        GameApiRequest request = new GameApiRequest();
        request.setEntryPoint(entryPoint);
        request.setBody(body);

        request.setUIResponseHandler(new Handler(){
            @Override
            public void handleMessage(Message msg) {

                    if(msg.obj != null){
                        if(msg.what == MCStatics.ResponseMessageCode)
                            callback.OnSuccess(msg.obj);
                        else
                            callback.OnFail("please,try again later.");

                    }   
                }

            }
        });

        Thread thread = new Thread(request);
        thread.start();
    }

Answer 1

我使用countdownlatch解决了这个问题，

private final CountDownLatch responseLatch = new CountDownLatch (1);

MRURLProtocol.mapDomain("gamification.fs", new MRProtocolResponseHandler() {
            @Override
            public InputStream handle(Uri uri) {
             InputStream stream = null;
                myApplication.getChamp().getImpulser().invokeRequest(method, body,new OnWebApiResponseArrived() {
                    @Override
                    public void OnSuccess(Object obj) {
                        String configString = "(function () { if (typeof(window) === 'undefined') ..... %s ....";
                        configString = String.format(configString, obj);
                        stream = new ByteArrayInputStream(configString.getBytes());

                        responseLatch.countDown();
                    }

                    @Override
                    public void OnFail(String errMsg) {
                        String configString = "(function () { if (typeof(window) === 'undefined')..... %s ....";
                        configString = String.format(configString, errMsg);
                        stream = new ByteArrayInputStream(configString.getBytes());

                        responseLatch.countDown();
                    }
                }); 

                try {
                    responseLatch.await();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }

                return stream;
            }
        });

如何在同步方法中使用新线程

1 个答案: