如何忽略错误并继续无限流? [英] How to ignore error and continue infinite stream?

问题描述

我想知道如何忽略异常并继续无限流(在我的情况下是位置流)?

I would like to know how to ignore exceptions and continue infinite stream (in my case stream of locations)?

我正在获取当前用户位置(使用 Android-ReactiveLocation )，然后将其发送到我的API(使用 Retrofit ).

I'm fetching current user position (using Android-ReactiveLocation) and then sending them to my API (using Retrofit).

在我的情况下，当在网络调用过程中发生异常(例如超时)时，将调用onError方法，流将自行停止.如何避免呢?

In my case, when exception occurs during network call (e.g. timeout) onError method is invoked and stream stops itself. How to avoid it?

活动:

private RestService mRestService;
private Subscription mSubscription;
private LocationRequest mLocationRequest = LocationRequest.create()
            .setPriority(LocationRequest.PRIORITY_HIGH_ACCURACY)
            .setInterval(100);
...
private void start() {
    mRestService = ...;
    ReactiveLocationProvider reactiveLocationProvider = new ReactiveLocationProvider(this);
    mSubscription = reactiveLocationProvider.getUpdatedLocation(mLocationRequest)
            .buffer(50)
            .flatMap(locations -> mRestService.postLocations(locations)) // can throw exception
            .subscribeOn(Schedulers.newThread())
            .observeOn(AndroidSchedulers.mainThread())
            .subscribe();
}

RestService:

RestService:

public interface RestService {
    @POST("/.../")
    Observable<Response> postLocations(@Body List<Location> locations);
}

推荐答案

mRestService.postLocations(locations)发出一项，然后完成. 如果发生错误，则会发出错误，从而完成流.

mRestService.postLocations(locations) emit one item, then complete. If an error occur, then it emit the error, which complete the stream.

在flatMap中调用此方法时，错误继续出现在主"流上，然后流停止.

As you call this method in a flatMap, the error continue to your "main" stream, and then your stream stops.

您所能做的就是将您的错误转换为另一个项目(如此处所述: https://stackoverflow.com/a/28971140/476690 )，但不在您的主流(如我想您已经尝试过)上，而是在mRestService.postLocations(locations)上.

What you can do is to transform your error into another item (as described here : https://stackoverflow.com/a/28971140/476690 ), but not on your main stream (as I presume you already tried) but on the mRestService.postLocations(locations).

这样，此调用将发出一个错误，该错误将转换为一个项/另一个可观察的项，然后完成. (不调用onError).

This way, this call will emit an error, that will be transformed to an item/another observable and then complete. (without calling onError).

在消费者视图中，mRestService.postLocations(locations)将发出一项，然后完成，就像一切都成功了.

On a consumer view, mRestService.postLocations(locations) will emit one item, then complete, like if everything succeed.

mSubscription = reactiveLocationProvider.getUpdatedLocation(mLocationRequest)
        .buffer(50)
        .flatMap(locations -> mRestService.postLocations(locations).onErrorReturn((e) -> Collections.emptyList()) // can't throw exception
        .subscribeOn(Schedulers.newThread())
        .observeOn(AndroidSchedulers.mainThread())
        .subscribe();

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