[英]convert rjxs map and flatten/reduce to flatMap
I believe the following code can be refactored using a flatMap but I cant seem to get it working as desired. 我相信以下代码可以使用flatMap进行重构,但是我似乎无法使其按需工作。
I understand flatMap essentially maps and then flattens, which is perfect for me as I'm using forkJoin so get an array of responses back from getAutocompleteSuggestions(). 我了解flatMap本质上是映射,然后进行展平,这对我来说是完美的,因为我正在使用forkJoin,因此可以从getAutocompleteSuggestions()返回一系列响应。
I want a single array of results upon subscription (which is what the code below produces), but changing the top level map to flatMap sends multiple single objects to the subscription. 我希望在订阅时得到一个结果数组(这是下面的代码生成的结果),但是将顶级映射更改为flatMap会将多个单个对象发送给订阅。 How can this code be better written with flatMap()? 如何使用flatMap()更好地编写此代码?
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 $subs
return Observable.forkJoin(...subs$);
}
It looks like there might be a bit of confusion between the RxJS flatMap
and the Array prototype method flatMap
. 看起来RxJS flatMap
与Array原型方法flatMap
之间可能有些混乱。 Note that the purpose of the RxJS flatMap is not to flatten arrays that are the subjects of the stream, but rather to flatten a stream of Obervables into a single observables. 请注意,RxJS flatMap的目的不是将流的主题数组展平,而是将Obervable的流展平为单个Observable。 See this SO post: 看到这样的帖子:
Why do we need to use flatMap? 为什么我们需要使用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. ...基本上,如果Observable表示一个推入类型T的值的可观察对象,则flatMap将类型T'-> Observable的函数作为其参数,并返回Observable。 map takes a function of type T' -> T and returns Observable. map采用类型T'-> T的函数并返回Observable。
If you want your code to be a bit cleaner, you could use the myArray.flatMap
method. 如果希望代码更myArray.flatMap
一点,可以使用myArray.flatMap
方法。 Here's a potential answer to your question with the Array flatMap
method: 这是使用Array flatMap
方法对您的问题的可能答案:
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 $subs return Observable.forkJoin(...subs$); }
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