swift - Rx swift : Zip Observables only if requirements are met

Question

想象一个用户配置文件，它允许编辑姓名、姓氏、年龄 和avatarImage 等字段。当用户点击 Save 时，系统会针对每个已更改的值发送请求。

现在请考虑具有如下签名的函数:

func rx_updateUserName(name: String) -> Observable<UpdateUserNameResponse>
func rx_updateSurname(surname: String) -> Observable<UpdateSurnameResponse>
func rx_updateAge(age: Int) -> Observable<UpdateAgeResponse>

通常我们会像这样压缩这些请求:

let nameReqObservable = rx_updateUserName("Gandalf")
let surnameReqObservable = rx_updateSurname("The White")
let ageReqObservable = rx_updateAge(123)

let zippedRequests = Observable.zip(nameReqObservable, surnameReqObservable, ageReqObservable, 
    resultSelector: { (userNameResponse, surnameResponse, areResponse) in

    return (userNameResponse, surnameResponse, areResponse)
}).subscribeNext(...)

只想对变化的值执行请求怎么办？

Answer 1

你试过.combineLatest()了吗？此处可视化 - http://rxmarbles.com/#combineLatest在您的例子中，只要其中一个输入可观察对象的值发生变化，combinedRequests 可观察对象就会发出一组新的输入。

let combinedRequests = Observable.combineLatest(nameReqObservable, surnameReqObservable, ageReqObservable, 
    resultSelector: { (userNameResponse, surnameResponse, ageResponse) in

    return (userNameResponse, surnameResponse, ageResponse)
}).subscribeNext(...)

来自 RxSwift 的文档，.combineLatest():

Merges the specified observable sequences into one observable sequence by using the selector function whenever any of the observable sequences produces an element.