rxSwift按钮显示观察者问题

时间:2017-08-29 13:52:59

标签: ios swift uibutton rx-swift

我有一个init,它在我的viewModel上为rxSwift设置观察者的处理

init(login: Observable<String>, password: Observable<String>, buttonPress: Observable<Void>) {
        let userInputs = Observable.combineLatest(login, password) { (login, password) -> (String, String) in
            return (login, password)
        }

        let callFromPress = buttonPress
            .withLatestFrom(userInputs)
            .map { (login, password) in
                self.makeLoginRequest(userLogin: login, loginPassword: password, loginSecret: self.loginSecret, deviceToken: self.deviceToken)
        }
    }

然而,当我点击按钮时,没有任何反应。它连接在VC中,因为:

buttonPress: self.loginView.loginButton.rx.tap.asObservable()

关于按钮敲击为什么没有触发任何错误的任何解决方案?

更新

init(login: Observable<String>, password: Observable<String>, buttonPress: Observable<Void>) {
    let userInputs = Observable.combineLatest(login, password) { (login, password) -> (String, String) in
        return (login, password)
    }

    buttonPress.subscribe(onNext: { each in
        print(each)
            .withLatestFrom(userInputs)
            .map { (login, password) in
                self.makeLoginRequest(userLogin: login, loginPassword: password, loginSecret: self.loginSecret, deviceToken: self.deviceToken)
        }
    }).disposed(by: self.disposeBag)
}

错误:

  

无关的参数标签'onNext:'在通话中

更新2:

init(login: Observable<String>, password: Observable<String>) {
    // each time the login and password change, returns login and string value
    didTapLoginButton = { [weak self] _ in
        // allows for strong self reference
        guard let `self` = self else { return }

        Observable.combineLatest(login, password) { (login, password) in
            //
            }
            .subscribe(onNext: { response in
                self.makeLoginRequest(userLogin: login, loginPassword: password, loginSecret: self.loginSecret, deviceToken: self.deviceToken)
                // do something with your response
            })
            .disposed(by: self.disposeBag)

    }
}

2 个答案:

答案 0 :(得分:2)

添加.subscribe。在具有订户之前,可观察者不会发送事件。

示例:

let observable = Observable.from([1,2,3,4,5,6]) // won't send events. It's only the declaration. 

当您订阅它时,事情就会发生:

observable.subscribe(onNext: { each in 
    print(each) // Will print each number
})
.disposed(by: disposeBag) // add a dispose bag 

我没有尝试过,但在你的特定情况下,我会做这样的事情:

1-在viewModel中,添加一个clousure属性:

var didTapLoginButton: () -> Void = { _ in }

2-在viewController - viewDidLoad()方法中:

self.loginView.loginButton.rx.tap.asObservable()
    .subscribe(onNext: { [weak self] _ in 
        self?.viewModel.didTapLoginButton()
    })
    .disposed(by: bag)

3-在viewModel

init(login: Observable<String>, password: Observable<String>) {

    didTapLoginButton = { [weak self] _ in 
        // This let you call self as strong reference instead optional
        guard let `self` = self else { return }

        Observable.combineLatest(login, password) { (login, password) -> (String, String) in
            self.makeLoginRequest(userLogin: login, loginPassword: password, loginSecret: self.loginSecret, deviceToken: self.deviceToken)
        }
        .subscribe(onNext: { response in 
            // do something with your response
        })
        .disposed(by: bag)

    }
}

答案 1 :(得分:1)

就我而言,我正在做

subscribe(onNext: { [weak self] in

我应该这样做

subscribe(onNext: { [weak self] _ in

这消除了错误