我正在尝试实现一个具有最大重试次数的alamofire调用。代码如下:
RxAlamofire.request(.post, URL, parameters: parameters, encoding: JSONEncoding.default)
.observeOn(MainScheduler.instance)
.retryWhen { (errors: Observable<Error>) in
return errors.flatMapWithIndex { (e, a) -> Observable<Int64> in
if a >= self.RETRY_COUNT - 1 {
return Observable.error(e)
}
print("Error: delay server call retry by \(a+1) second(s)")
return Observable<Int64>.timer(RxTimeInterval(a+1), scheduler: MainScheduler.instance)
}
}
.subscribe(
onNext: {
(result) in
print("I get here when retrying...")
},
onError: { (error) in
print(error)
}
)
.addDisposableTo(self.disposeBag)不幸的是,在重试时,我进入了subscribe中的onNext块-我不想在得到结果之前到达那里。(如预期的那样,超过最大重试次数后,onError会给出错误)。有人能帮帮忙吗?
发布于 2018-03-16 16:11:37
我试图重现您的场景,但以下代码不会在出现错误时触发订阅中的onNext闭包。我稍微重写了你的代码,我的例子总是出错。我使用的是RxSwift 4.0.0。
let count = 2
enum MyError: Error {
case crash
}
_ = Observable<Int>.error(MyError.crash)
.debug("prior")
.retryWhen { errors in
return errors.enumerated().flatMap { (index, error) -> Observable<Void> in
guard index < count - 1 else { return .error(error) }
print("Error: delay server call retry by \(index + 1) second(s)")
return Observable<Void>.just(()).delay(RxTimeInterval(index + 1), scheduler: MainScheduler.instance)
}
}
.debug("after")
.subscribe(onNext: { element in
print("got next element: \(element)")
})这将产生以下输出。
2018-03-16 09:05:16.921: after -> subscribed 2018-03-16 09:05:16.924: prior -> subscribed 2018-03-16 09:05:16.924: prior -> Event error(blok) Error: delay server call retry by 1 second(s) 2018-03-16 09:05:16.925: prior -> isDisposed 2018-03-16 09:05:17.926: prior -> subscribed 2018-03-16 09:05:17.926: prior -> Event error(blok) 2018-03-16 09:05:17.927: after -> Event error(blok) 2018-03-16 09:05:17.928: after -> isDisposed 2018-03-16 09:05:17.928: prior -> isDisposed
https://stackoverflow.com/questions/42271750
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