创建的可观察对象不发出任何东西

Question

我正在尝试创建一个 returns Observable<(HTTPURLResponse, Any)> 使用 RxAlamofire:

的函数

class APIManager {
    let disposeBag = DisposeBag()
    func getResponse(credentialData: Credentials, ulr: String)->Observable<(HTTPURLResponse, Any)>{
    let credentialData = "\(credentialData.username):\(credentialData.password)".data(using: String.Encoding(rawValue: String.Encoding.utf8.rawValue))!
    let base64Credentials = credentialData.base64EncodedString()
    let headers = ("Authorization", "Basic \(base64Credentials)")
    let header = HTTPHeaders.init(dictionaryLiteral: headers)
    return Observable.create{ observer in
        requestJSON(.get, ulr, parameters: nil, encoding: JSONEncoding.default, headers: header)
        .subscribe(onNext: { response in
            observer.onNext(response)
            } ,


    onError: { error in
        observer.onError(error)
        },
    onCompleted: {},
    onDisposed: {}).disposed(by: self.disposeBag)
    return Disposables.create()
    }
}

}

进入onNext的代码没有执行；它仅通过 onDisposed。

我该如何解决？

Answer 1

为什么你用另一个 Observable 包装 requestJSON 函数，它只是转发相同的元素和错误，而那个函数已经返回你需要的 Observable？

这还不够吗？

class APIManager {
    func getResponse(credentialData: Credentials, ulr: String) -> Observable<(HTTPURLResponse, Any)> {
        let credentialData = "\(credentialData.username):\(credentialData.password)"
            .data(using: String.Encoding(rawValue: String.Encoding.utf8.rawValue))!
        let base64Credentials = credentialData.base64EncodedString()
        let headers = ("Authorization", "Basic \(base64Credentials)")
        let header = HTTPHeaders.init(dictionaryLiteral: headers)
        return requestJSON(.get, ulr, parameters: nil, encoding: JSONEncoding.default, headers: header)
    }
}

创建的可观察对象不发出任何东西

Created observable doesn't emit anything

swift

rx-swift

rxalamofire