如果getSearchResult(keyword)返回一个对象数组，则可以使用Array.flat()将嵌套数组展平
但不是很RXJS。
这是一种Rx方式

of(keywordsArray)      // observable stream from keywords
.pipe(
  concatMap(keyword => {          // convert each keyword to search results
    return getSearchResult(keyword)
  })
)
.subscribe(searchResult => ...)

或

const searchResults$ = of(keywordsArray)      // observable stream from keywords
.pipe(
  concatMap(keyword => {          // convert each keyword to search results
    return getSearchResult(keyword)
  })
)

您可以使用Array.map()来迭代关键字，并使用concat()来合并它们：

const resultObservable: Observable<searchObject[]>[] = keywordsArray.map(keyword => 
    getSearchResult(keyword)
);

concat(resultObservable).subscribe(() => { /* Do Some Stuff */ })

您应该使用flatMap()或替代方案：

const resultObservable: Observable<searchObject[]> [] = keywordsArray.map(keyword => {
  return getSearchResult(keyword);
})
.reduce((x, y) => x.concat(y), []);

您可以使用forkJoin & of from RxJS和Array.reduce来实现，如下所示：

const resultObservable = keywordsArray.map((keyword) => {
  return of(getSearchResult(keyword)); //function returns an array of searchObject
});

// forkJoin combines the the array of observables and returns Observable<searchObject[][]>
forkJoin(resultObservable)
  .pipe(
    map((res) =>
      // Array.reduce will combine the arrays and return one array that has all the objects.
      res.reduce((acc, curr) => [...acc, ...curr], [])
    )
  )
  .subscribe((res: searchObject[]) => {
    // Do here what you need with the combined result
  });