如何合并两个 Observable。处方 6.2.2

Question

我有两个后端调用，只有在两个响应都准备就绪时我才需要处理结果：

this.historyService.fetchBalance().subscribe((balance: Balance[]) => {
    this.servicesBalance = balance;
});

this.historyService.fetchOuts().subscribe((outs: History[]) => {
    this.outHistories = outs;
});

我知道 forkJoin 方法，但它 returns 数组，我丢失了类型，像这样：

...subscribe([v1, v2] => ...)

我想像这样保存返回值的类型：

...subscribe(balance: Balance[], outs: History[] => ...)

在我的案例中如何合并 Observables？

myQuestionUnionMethod(
    this.historyService.fetchBalance(),
    this.historyService.fetchOuts()
)
    .subscribe(balance: Balance[], outs: History[] => ...)

Answer 1

你试过zip了吗？好像是properly typed:

export function zip<T, T2>(v1: ObservableInput<T>, v2: ObservableInput<T2>): Observable<[T, T2]>;
export function zip<T, T2, T3>(v1: ObservableInput<T>, v2: ObservableInput<T2>, v3: ObservableInput<T3>): Observable<[T, T2, T3]>;
export function zip<T, T2, T3, T4>(v1: ObservableInput<T>, v2: ObservableInput<T2>, v3: ObservableInput<T3>, v4: ObservableInput<T4>): Observable<[T, T2, T3, T4]>;
export function zip<T, T2, T3, T4, T5>(v1: ObservableInput<T>, v2: ObservableInput<T2>, v3: ObservableInput<T3>, v4: ObservableInput<T4>, v5: ObservableInput<T5>): Observable<[T, T2, T3, T4, T5]>;
export function zip<T, T2, T3, T4, T5, T6>(v1: ObservableInput<T>, v2: ObservableInput<T2>, v3: ObservableInput<T3>, v4: ObservableInput<T4>, v5: ObservableInput<T5>, v6: ObservableInput<T6>): Observable<[T, T2, T3, T4, T5, T6]>;

再想想，forkJoin 似乎也输入正确：

export function forkJoin<T>(sources: [ObservableInput<T>]): Observable<T[]>;
export function forkJoin<T, T2>(sources: [ObservableInput<T>, ObservableInput<T2>]): Observable<[T, T2]>;
export function forkJoin<T, T2, T3>(sources: [ObservableInput<T>, ObservableInput<T2>, ObservableInput<T3>]): Observable<[T, T2, T3]>;
export function forkJoin<T, T2, T3, T4>(sources: [ObservableInput<T>, ObservableInput<T2>, ObservableInput<T3>, ObservableInput<T4>]): Observable<[T, T2, T3, T4]>;
export function forkJoin<T, T2, T3, T4, T5>(sources: [ObservableInput<T>, ObservableInput<T2>, ObservableInput<T3>, ObservableInput<T4>, ObservableInput<T5>]): Observable<[T, T2, T3, T4, T5]>;
export function forkJoin<T, T2, T3, T4, T5, T6>(sources: [ObservableInput<T>, ObservableInput<T2>, ObservableInput<T3>, ObservableInput<T4>, ObservableInput<T5>, ObservableInput<T6>]): Observable<[T, T2, T3, T4, T5, T6]>;
export function forkJoin<T>(sources: Array<ObservableInput<T>>): Observable<T[]>;

为什么您认为 return 值输入不正确？你有例子吗 "lost types"?

如何合并两个 Observable。处方 6.2.2

How to union two Observable. Rx 6.2.2

typescript

rxjs6