Question

我需要知道引发错误后如何保留订阅。

我创建了这个可观察的对象：

let saveClick$ = Observable.fromEvent(this.saveButton.nativeElement, 'click');

因此，我尝试使用Observable.empty，但据我所知，它发出了一个完整的消息，因此将删除订阅者：

this.saved$ = saveClick$.pipe(this.pushUser(), catchError(() => Observable.empty<AdministrationUser>()), share());

我希望在遇到任何错误时都保留订阅。

有什么想法吗？

其他代码：

// Custom pipes
private push = <T>() => switchMap<T, AdministrationUser>(() => this.service.push(this.user));
private handleError = <T>() => catchError<T, Array<{code: string, message: string}>>((error: ResponseError) => Observable.of(error.errors));
private handleEmptyUser = <T>() => catchError<T, AdministrationUser>(() => Observable.of(UsuarisadministracioSubcomponentComponent.EMPTY_USER));

private pushUser = () => pipe( //simplified pipe
    this.push()
);

我的服务是：

const buildURL = () => map((filter: {userId: string} & Trace) => this.buildPushURL(filter.currentUser, filter.currentApplication, filter.userId));
const makeRequest = () => switchMap((url: string) => httpMethodFn.call(this.authHttp, url, user));
const buildResponse = () => map(() => user);
const onErrorGetDetails = () => catchError((error: Response) => Observable.throw(<ResponseError>error.json()));

return Observable.of({userId: user.id})
    .pipe(
        buildURL(),
        makeRequest(),
        buildResponse(),
        onErrorGetDetails()
    );

Answer 1

您应该在内部可观察的管道中而不是在源可观察的地方捕获错误。如果您在可观察到的源上捕获错误，那么按照rxjs概念，可观察到的源在发生错误后将不会发出新值。因此，像这样在内部可观察的地方捕获错误：

this.saved$ = saveClick$.pipe(
                               //i am assuming this.pushUser() is a higher order function which returns an observable
                               this.pushUser()
                                   .pipe(
                                          catchError(() => Observable.empty<AdministrationUser>())
                                        ),
                                        share()
                                    );

在内部可观察范围内捕获错误将使您的源可观察活动保持活动状态，即使内部可观察范围引发错误，也将继续发出值。您可以根据需要移动share()运算符[即不论是在内部可观测管道内还是在源可观测管道内]，但想法仍然相同-在内部可观测管道内捕获错误，以使源（外部）可观测活动保持活跃。

编辑1：[根据用户的最新代码建议简化代码]

请在服务中定义您的push方法，如下所示：

push(user) {

    //By seeing your code; I could not figure out where and how are you passing the parameter in buildUrl
    //It appears to me that this.buildPushURL method simply return an URL based on the passed parameters and IT DOES NOT MAKE
    //ANY HTTP CALL TO YOUR BACKEND. IF THAT IS TRUE - 
    //THEN please adjust the below code accordingly as I dont know how you use this method OR PLEASE provide some more detaling 
    //on  this.buildPushURL() method
    const url = this.buildPushURL(filter.currentUser, filter.currentApplication, filter.userId);

    //I am assuming that httpMethodFn calls http.get() or http.post() and returns an observable
    //Suggestion - Why you dont use httpClient.get or httpClient.post() directly? It will avoid to call
    //httpMethodFn by httpMethodFn.call?

    //Bottom line is - httpMethodFn.call must return an observable otherwise adjust your httpMethodFn.call code to return an observable
    //to make below code work.
    return httpMethodFn.call(this.authHttp, url, user)
                       .pipe(
                           map(user => {
                               console.log(user);
                               return user;
                           }),
                           catchError(error => throwError(<ResponseError>error.json()))
                       );

}

并像这样使用它-

yourMethodReturnsAnObservableWhichIsSubscribeByTheConsumer() {

    const saveClick$ = Observable.fromEvent(this.saveButton.nativeElement, 'click');

    return saveClick$.pipe(

            switchMap(() => {
                return this.push(user)
                            .pipe(
                                //you catch error here so that your outer observable will keep alive

                                //please adjust you code as per your need
                                //I guess it will give you an idea
                                handleError()
                            )
            })


    );
}

希望您将了解如何简化代码。由于我不了解您的设计和完整的代码，因此我无法提供完美的解决方案。我很确定上面的代码会给您一个想法。

一个建议-最好使用高阶函数，但要理解不要过度使用它们，因为这会使您的代码有点难以理解[PS-这是我个人的观点，并且有不同的观点是很好的。。:)]。通过应用编码的功能样式，使可观察链尽可能简单。 RXJS很棒：）

Answer 2

您需要使用catchError运算符并返回NEVER Observable

 import { NEVER } from 'rxjs';
 import { catchError } from 'rxjs/operators';

 observable$.pipe(catchError((error: any) => {return NEVER;}));

Angular rxjs：在出现错误后保留订阅

2 个答案: