Question

我有下面这段代码，我希望我的线程等待，直到调用任何一个回调函数。

发出我的代码点击我正在进行synchObj等待的行，但之后它就消失了它不会移动到任何地方。

如果在等待它之前不会如何调用notify？

iGPlaceApi.getStreams(new Callback<mGooglePlacesApiResponse>() {

    @Override
    public void failure(RetrofitError retrofitError) {
        String serverResponse = retrofitError.toString();
        synchronized (synchObj) {
            synchObj.notify();
            //synchObj.notifyAll();
        }
    }

    @Override
    public void success(mGooglePlacesApiResponse googlePlacesObj, Response arg1){
        nearbyPlaces = new String[googlePlacesObj.results.size()][4];
        for (int i = 0; i < googlePlacesObj.results.size(); i++) {
            mGooglePlaces.place place = googlePlacesObj.results.get(i);
            nearbyPlaces[i][0] = place.icon;
            nearbyPlaces[i][1] = place.name;
            nearbyPlaces[i][2] = String.valueOf(place.geometry.location.lat);
            nearbyPlaces[i][3] = String.valueOf(place.geometry.location.lng);
        }
        synchronized (synchObj) {
            synchObj.notify();
            //synchObj.notifyAll();
        }
    }

});

synchronized (synchObj) {
    synchObj.wait();
}

Handler handler=new Handler();
            Runnable thr = new Runnable() {
                public void run() {
                    iGPlaceApi.getStreams(new Callback<mGooglePlacesApiResponse>() {

                        @Override
                        public void failure(RetrofitError retrofitError) {
                            String serverResponse = retrofitError.toString();
                            synchronized (synchObj) {
                                synchObj.notifyAll();
                            }
                        }

                        @Override
                        public void success(mGooglePlacesApiResponse googlePlacesObj, Response arg1) {
                            nearbyPlaces = new String[googlePlacesObj.results.size()][4];
                            for (int i = 0; i < googlePlacesObj.results.size(); i++) {
                                mGooglePlaces.place place = googlePlacesObj.results.get(i);
                                nearbyPlaces[i][0] = place.icon;
                                nearbyPlaces[i][1] = place.name;
                                nearbyPlaces[i][2] = String.valueOf(place.geometry.location.lat);
                                nearbyPlaces[i][3] = String.valueOf(place.geometry.location.lng);
                            }
                            synchronized (synchObj) {
                                synchObj.notifyAll();
                            }

                        }

                    });
                }
            };
            handler.post(thr);

            synchronized (synchObj) {
                synchObj.wait();
            }

Answer 1

你不能这样做。回调将在调用getStreams方法的同一线程上调用。

在调用方法返回之前，无法调用回调。您可能需要在另一个线程中调用getStreams。

Answer 2

你做错了。处理程序任务将在同一个线程上执行，因此在您的情况下它将永远不会被执行。如果你想等待你的任务完成，你应该像这样使用Thread：

// defined somewhere
boolean done = false;
Thread thr=new Thread(Runnable() {
    public void run() {
        iGPlaceApi.getStreams(new Callback<mGooglePlacesApiResponse>() {
            @Override
            public void failure(RetrofitError retrofitError) {
                String serverResponse = retrofitError.toString();
                        synchronized (synchObj) {
                            done = true;
                            synchObj.notifyAll();
                        }
                    }

                    @Override
                    public void success(mGooglePlacesApiResponse googlePlacesObj, Response arg1) {
                        nearbyPlaces = new String[googlePlacesObj.results.size()][4];
                        for (int i = 0; i < googlePlacesObj.results.size(); i++) {
                            mGooglePlaces.place place = googlePlacesObj.results.get(i);
                            nearbyPlaces[i][0] = place.icon;
                            nearbyPlaces[i][1] = place.name;
                            nearbyPlaces[i][2] = String.valueOf(place.geometry.location.lat);
                            nearbyPlaces[i][3] = String.valueOf(place.geometry.location.lng);
                        }
                        synchronized (synchObj) {
                            done = true;
                            synchObj.notifyAll();
                        }

                    }

                });
            }
        });
        thr.start();

        synchronized (synchObj) {
            while (!done) {
                synchObj.wait();
            }
        }

请注意，我不是只使用runnable，而是使用Thread，并且指示任务状态是必需的，因为等待调用可以自动唤醒，如果任务还没有完成，你必须再次调用它。

调用等待后线程不移动以通知

2 个答案: