Question

在更改＆＃34; SortBy＆＃34;上，我的程序将执行NetworkIO以检索热门电影并显示它们。

但是，似乎虽然我已完成subscribeOn(Schedulers.io())，但MovieDB.getPopular()中的MovieDB.getTopRated()函数中的NetworkIO call和map已在主要内容中执行线程，我得到android.os.NetworkOnMainThreadException。

我想知道如何使public Movie[] call(SortBy sortBy)异步。

sortObservable.map(new Func1<SortBy, Movie[]>() {
    @Override
    public Movie[] call(SortBy sortBy) {
        try {
            switch (sortBy) {
                case POPULAR:
                    return MovieDB.getPopular(); // NETWORK IO
                case TOP_RATED:
                    return MovieDB.getTopRated(); // NETWORK IO
            }
        } catch (IOException e) {
            e.printStackTrace();
        } catch (JSONException e) {
            e.printStackTrace();
        }
        return new Movie[0];
    }
})
        .subscribeOn(Schedulers.io())
        .observeOn(AndroidSchedulers.mainThread())
        .subscribe(new Action1<Movie[]>() {
            @Override
            public void call(Movie[] movies) {
                imageAdapter.loadData(movies);
            }
        });

Answer 1

最后，我自己解决了这个问题：

sortObservable.flatMap(new Func1<SortBy, Observable<Movie[]>>() {

    @Override
    public Observable<Movie[]> call(SortBy sortBy) {
        switch (sortBy) {
            case POPULAR:
                return Observable.fromCallable(() -> MovieDB.getPopular()).subscribeOn(Schedulers.io());
            case TOP_RATED:
                return Observable.fromCallable(() -> MovieDB.getTopRated()).subscribeOn(Schedulers.io());
            default:
                return Observable.fromCallable(() -> new Movie[0]).subscribeOn(Schedulers.io());
        }
    }
})
        .observeOn(AndroidSchedulers.mainThread())
        .subscribe(new Action1<Movie[]>() {
            @Override
            public void call(Movie[] movies) {
                imageAdapter.loadData(movies);
            }
        });

Answer 2

请检查以下内容是否适合您。它使用flatMap代替map。

sortObservable.flatMap(new Func1<SortBy, Observable<Movie[]>>() {

        @Override
        public Observable<Movie[]> call(SortBy sortBy) {
            try {
                switch (sortBy) {
                    case POPULAR:
                        return Observable.just(MovieDB.getPopular()); // NETWORK IO
                    case TOP_RATED:
                        return Observable.just(MovieDB.getTopRated()); // NETWORK IO
                }
            } catch (IOException e) {
                e.printStackTrace();
            } catch (JSONException e) {
                e.printStackTrace();
            }
            return Observable.just(new Movie[0]);
        }
    }).subscribe(new Action1<Movie[]>() {
        @Override
        public void call(Movie[] movies) {
            imageAdapter.loadData(movies);
        }
    });

从Github上的源代码中，您似乎正在使用OkHttp执行请求的同步模式。 OkHttp还支持异步请求，这可能是首选。以下是几种方法所需的变化。

run方法应使用enqueue代替execute。

Observable<String> runAsync(String url){
return Observable.create(subscriber -> {
    Request request = new Request.Builder().url(url).build();

    client.newCall(request).enqueue(new Callback() {

        @Override
        public void onResponse(Call call, Response response) throws IOException {
            subscriber.onNext(response.body().string());
        }

        @Override
        public void onFailure(Call call, IOException e) {
            subscriber.onError(e);
        }
    });
});
}

getApi可以返回Observable<Movie[]>而不是Movie[]

public Observable<Movie[]> getApiAsync(String type){
return runAsync("http://api.themoviedb.org/3/movie/" + type
        + "?api_key=412e9780d02673b7599233b1636a0f0e").flatMap(response -> {
            Gson gson = new Gson();
            Map<String, Object> map = gson.fromJson(response,
                    new TypeToken<Map<String, Object>>() {
                    }.getType());
            Movie[] movies = gson.fromJson(gson.toJson(map.get("results")),
                    Movie[].class);
            return Observable.just(movies);
        });
}

Android RxJava在map函数中异步调用

2 个答案: