为了简化问题,我在这里使用了数字和字符串。代码:
const numbers$:Observable<number[]> = of([1,2,3]);
const strings: string[] = ["a","b"];
function getStrings(): Observable<string>[]{
return numbers$.pipe(
map((numbers: number[]) => {
const strings$: Observable<string>[] = strings.map(s => of(s));
return strings$;
}),
)
}
getStrings().subscribe(x => console.log(x))
我得到的错误是:
Type 'Observable<Observable<string>[]>' is missing the following properties from type 'Observable<string>[]
如何从Observable<string>[]
函数获得getStrings
?我尝试使用flatMap,switchMap,但无法获得完美的组合。
答案 0 :(得分:1)
您需要使用mergeMap()
和forkJoin()
:
function getStrings(): Observable<string[]>{
return numbers$.pipe(
mergeMap((numbers: number[]) => {
const strings$: Observable<string>[] = strings.map(s => of(s));
return forkJoin(strings$);
}),
)
}
getStrings().subscribe(x => console.log(x))
答案 1 :(得分:0)
据我了解,您想通过可观察的方式将两个列表“压缩”在一起?
我可以给你这个
const numbers$: Observable<number[]> = of([1, 2, 3]);
const strings$: Observable<string[]> = of(['a', 'b']);
const combined$: Observable<any> = zip(numbers$, strings$)
.pipe(
map(([numbers, strings]) =>
numbers.length > strings.length ?
numbers.map((value, index) => [value, strings[index]]) :
strings.map((value, index) => [numbers[index], value]))
);
combined$.subscribe(value => console.log(value));
这将记录: [ [ 1, “一种” ], [ 2, “ b” ], [ 3, 空值 ] ]