假设我有可观察的class PaginationNetworkModel<T1: Mappable>: NSObject {
let refreshTrigger = PublishSubject<Void>()
let loadNextPageTrigger = PublishSubject<Void>()
let loading = Variable<Bool>(false)
let elements = Variable<[T1]>([])
var offset:Int = 0
let error = PublishSubject<Swift.Error>()
private let disposeBag = DisposeBag()
override init() {
super.init()
let refreshRequest = loading.asObservable()
.sample(refreshTrigger)
.flatMap { loading -> Observable<Int> in
if loading {
return Observable.empty()
} else {
return Observable<Int>.create { observer in
observer.onNext(0)
observer.onCompleted()
return Disposables.create()
}
}
}
let nextPageRequest = loading.asObservable()
.sample(loadNextPageTrigger)
.flatMap { [unowned self] loading -> Observable<Int> in
if loading {
return Observable.empty()
} else {
return Observable<Int>.create { [unowned self] observer in
self.offset += 1
observer.onNext(self.offset)
observer.onCompleted()
return Disposables.create()
}
}
}
let request = Observable
.of(refreshRequest, nextPageRequest)
.merge()
.shareReplay(1)
let response = request.flatMap { offset -> Observable<[T1]> in
self.loadData(offset: offset)
.do(onError: { [weak self] error in
self?.error.onNext(error)
}).catchError({ error -> Observable<[T1]> in
Observable.empty()
})
}.shareReplay(1)
Observable
.combineLatest(request, response, elements.asObservable()) { [unowned self] request, response, elements in
return self.offset == 0 ? response : elements + response
}
.sample(response)
.bind(to: elements)
.addDisposableTo(rx_disposeBag)
Observable
.of(request.map{_ in true},
response.map { $0.count == 0 },
error.map { _ in false })
.merge()
.bind(to: loading)
.addDisposableTo(rx_disposeBag)
}
func loadData(offset: Int) -> Observable<[T1]> {
return Observable.empty()
}
，我正在尝试创建可观察的A，它会发出两个事件：B发出事件时的第一个事件，以及5秒后的第二个事件。< / p>
到目前为止，我有以下内容：
A这很有效，但我觉得从一个闭包中订阅self.B = Observable.create { [unowned self] observer in
self.A.subscribe(onNext: {
observer.onNext(0)
self.delay(5) {
observer.onNext(1)
}
})
return Disposables.create()
}
是不可饶恕的。有没有更好的方法呢？
谢谢！
答案 0 :(得分：1)
解决方案是为延迟的observable重用a observable。下面是代码，以及概念证明。
let a = button.rx.tap.asObservable()
let delay = a.delay(5.0, scheduler: MainScheduler.instance)
let b = Observable.of(a, delay).merge()
b.subscribe(onNext: {
print("foo")
}).disposed(by: bag)