ios - rxSwift 按钮按下观察器问题

Question

我有一个 init 可以在我的 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)
        }
    }

但是当我点击按钮时，没有任何反应。它在 VC 中是这样连接的:

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

关于为什么点击按钮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:

Extraneous argument label 'onNext:' in call

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)

    }
}

Answer 1

添加.subscribe。在有订阅者之前，可观察对象不会发送事件。

例子:

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，但在您的特定情况下，我会这样做:

1- 在您的 viewModel 中，添加一个 clousure 属性:

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

2- 在您的 viewController - viewDidLoad() 方法中:

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

3- 在您的 viewModel 中:

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)

    }
}