开始之前首先开启Playground的needsIndefiniteExecution，以保证我们之后的延时操作能够正常运行，在你的Playground添加以下代码
import PlaygroundSupport
PlaygroundPage.current.needsIndefiniteExecution = true
retry
先来看看
retry，可能这个操作会比较常用，一般用在网络请求失败时，再去进行请求。
retry 就是失败了再尝试，重新订阅一次序列。
example(of: "retry") {
var count = 1
let funnyLookingSequence = Observable<Int>.create({ (observer) -> Disposable in
let error = NSError(domain: "test", code: 0, userInfo: nil)
observer.onNext(1)
observer.onNext(2)
observer.onNext(3)
if count < 2 {
observer.onError(error)
count += 1
}
observer.onNext(4)
observer.onNext(5)
observer.onNext(6)
return Disposables.create()
})
funnyLookingSequence.retry(2).subscribe({
print($0)
})
}
--- Example of: retry ---
next(1)
next(2)
next(3)
next(1)
next(2)
next(3)
next(4)
next(5)
next(6)
需要注意的是不加参数的 retry 会无限尝试下去。我们还可以传递一个 Int 值，来说明最多尝试几次。像这样 retry(2) ，最多尝试两次。
cacheError
出现 Error 时，用一个新的序列替换。
example(of: "cacheError") {
let sequenceThatFails = PublishSubject<Int>()
let recoverySequence = Observable.of(100,200,300)
sequenceThatFails.catchError({ (error) -> Observable<Int> in
return recoverySequence
}).subscribe({
print($0)
})
sequenceThatFails.onNext(1)
sequenceThatFails.onNext(2)
sequenceThatFails.onNext(3)
sequenceThatFails.onError(NSError(domain: "test", code: 2, userInfo: nil))
sequenceThatFails.onNext(4)
}
--- Example of: cacheError ---
next(1)
next(2)
next(3)
next(100)
next(200)
next(300)
completed
catchError 中你需要传递一个返回序列的闭包。
catchErrorJustReturn
catchErrorJustReturn遇到错误，返回一个值替换这个错误。
example(of: "catchErrorJustReturn") {
let sequenceThatFails = PublishSubject<Int>()
sequenceThatFails.catchErrorJustReturn(100).subscribe({
print($0)
})
sequenceThatFails.onNext(1)
sequenceThatFails.onNext(2)
sequenceThatFails.onNext(3)
sequenceThatFails.onError(NSError(domain: "test", code: 2, userInfo: nil))
}
--- Example of: catchErrorJustReturn ---
next(1)
next(2)
next(3)
next(100)
completed
