UI阻塞发生在Android中，尽管RxJava [英] UI blocking occurs in Android despite RxJava

问题描述

我使用RxJava移动网络接入到Android的一个单独的线程，但我的UI还是块。

I am using RxJava to move network access to a separate thread in Android, but my UI still blocks.

我没有使用了错误的观察到的，如下所示：的Andr​​oid RxJava，非阻塞

I am not using the wrong observable as shown here: Android RxJava, Non Blocking?

在codepoints [A]，[B]和[C]在下面code的传递顺序[A] - > [C] - > [B]所以目前螺纹加工精和RxJava调用[C]，一旦有了结果。这是好的。

The codepoints [A], [B] and [C] in below code are passed in the order [A] -> [C] -> [B] so the current thread is processed fine and RxJava calls [C] once it had a result. This is fine.

此外，阻塞好得多相比，这样做的UI线程上的网络电话，但我仍然有轻微阻塞。电话就被打出后的UI保持流畅，但是如果服务器未在几毫秒内做出反应，它会阻止。

Also, blocking is much better compared to doing the network call on the UI thread, but I still have minor blocking. The UI stays fluent after the call is made, but if the server does not respond in a matter of milliseconds, it blocks.

    private search; // search is an instance variable in the same class

    // [A]

    Observable.just(search.find("something")) // search.find calls the REST endpoint
            .subscribeOn(Schedulers.io()).observeOn(AndroidSchedulers.mainThread())
            .subscribe(new Action1<Search>() {

                @Override public void call(Search search) {
                    // further processing // [B]
                }

            }, new Action1<Throwable>() {

                @Override public void call(Throwable throwable) {
                    // error handler
                }

            });

    // [C]

这可能是一个问题，搜索是在可观察使用它的同一个类的实例变量，但终端呼叫从一个单独的库进行？它不应该的问题，对吧？

Could it be a problem that search is an instance variable in the same class where the Observable uses it, but the endpoint call is performed from a separate library? It shouldn't matter, right?

我在做什么不好，我不应该做？

Am I doing anything bad that I shouldn't be doing?

-

查找看起来是这样的（除去异常处理简洁）：

Find looks like this (removed exception handling for brevity):

public Search find(String searchtext) {
    setSearchtext(searchtext);
    SearchEndpoint.find(Session.getUser().getId(), searchtext);
    return this;
}

SearchEndpoint是这样的：

SearchEndpoint like this:

public static Search find(final Long userId, final String searchtext) throws IOException {
    return ApiService.api().searches().find(userId).setFind(searchtext).execute();
}

和将调用生成的谷歌云终端库。

and makes a call to the generated Google cloud endpoint library.

推荐答案

试试这个：

Observable.create(new Observable.OnSubscribe<Search>() {

    @Override
    // method signature is from memory - I hope I am correct...
    public void call(Subscriber<? super Search> subscriber) {
        try {
            Search search = search.find("something");
            subscriber.onNext(search);
            subscriber.onCompleted();
        } catch (SomeException e) {
            subscriber.onError(e);
        }
    }
})
// and then continue with your .subscribeOn(...)

要澄清一下，也许这让你的code中的问题更加明显：

To clarify, maybe this makes the problem with your code more obvious:

Observable.just(search.find("something"))

显然等同于

Search search = search.find("something");
Observable.just(search)

和这使得它很明显，search.find执行之前我们曾经交出了控制权交给rxjava，它是在执行任何线程的是当前 - 那么可观察到的从pre计算的价值建设和值的传递发生在另一个线程，但它不能帮助你多少...

And this makes it obvious that search.find is executed before we ever hand the control over to rxjava and it is executed on whatever thread you are currently on - then the construction of an Observable from the pre-computed value and the delivery of the value happen on another thread but that does not help you much...

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