我相信可以使用 flatMap 重构以下代码,但我似乎无法让它按预期工作。
我理解 flatMap 本质上是映射然后展平,这对我来说是完美的,因为我正在使用 forkJoin,因此可以从 getAutocompleteSuggestions() 返回一组响应。
我想要订阅时的单个结果数组(这是下面的代码生成的结果),但是将顶级映射更改为 flatMap 会向订阅发送多个单个对象。如何使用 flatMap() 更好地编写这段代码?
const $resultsObservable: Observable<any> = Observable.of(query)
.switchMap(q => this.getAutocompleteSuggestions(q))
//tried changing the below to flatMap()
.map(res => {
return res.map(resp => {
const content = resp.content;
content.map(result => this.someMethod(result));
return content;
})
.reduce((flatArr, subArray) => flatArr.concat(subArray), []);
});
getAutocompleteSuggestions(query: string): Observable<any> {
const subs$ = [];
//... add some observables to $subs
return Observable.forkJoin(...subs$);
}
最佳答案
看起来 RxJS flatMap
和数组原型(prototype)方法 flatMap
之间可能存在一些混淆。请注意,RxJS flatMap 的目的不是展平作为流主题的数组,而是将 Obervables 流展平为单个 observables。请参阅此帖子:
Why do we need to use flatMap?
... Basically if Observable denotes an observable object which pushes values of type T, then flatMap takes a function of type T' -> Observable as its argument, and returns Observable. map takes a function of type T' -> T and returns Observable.
如果您希望代码更简洁,可以使用 myArray.flatMap
方法。以下是使用 Array flatMap
方法对您的问题的潜在答案:
const $resultsObservable: Observable<any> = Observable.of(query)
.switchMap(q => this.getAutocompleteSuggestions(q))
// Flatmap on the array here because response is an array of arrays.
// The RxJS flatMap does not deal with flattening arrays
.map(res => res.flatMap(resp => {
const content = resp.content;
content.map(result => this.someMethod(result));
return content;
}));
getAutocompleteSuggestions(query: string): Observable < any > {
const subs$ = [];
//... add some observables to $subs
return Observable.forkJoin(...subs$);
}
关于javascript - 转换 rjxs map 并展平/缩减为 flatMap,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/49191962/