RxJava 和基于游标的 RESTful 分页

我正在使用 Spotify API，并希望使用 RxJava 链接一些分页结果。 Spotify 使用基于光标的分页，因此解决方案如下来自@lopar 的一个 https://stackoverflow.com/a/29594194/377260不管用。

回复来自这个电话 https://developer.spotify.com/web-api/get-followed-artists/看起来像这样（想象一下有 50 个items):

{
  "artists" : {
    "items" : [ {
      "id" : "6liAMWkVf5LH7YR9yfFy1Y",
      "name" : "Portishead",
      "type" : "artist"
    }],
    "next" : "https://api.spotify.com/v1/me/following?type=artist&after=6liAMWkVf5LH7YR9yfFy1Y&limit=50",
    "total" : 119,
    "cursors" : {
      "after" : "6liAMWkVf5LH7YR9yfFy1Y"
    },
    "limit" : 50,
    "href" : "https://api.spotify.com/v1/me/following?type=artist&limit=50"
  }
}

现在，我使用改造得到了前 50 个这样的结果：

public class CursorPager<T> {
    public String href;
    public List<T> items;
    public int limit;
    public String next;
    public Cursor cursors;
    public int total;

    public CursorPager() {
    }
}

public class ArtistsCursorPager {
    public CursorPager<Artist> artists;

    public ArtistsCursorPager() {
    }
}

then

public interface SpotifyService  {

    @GET("/me/following?type=artist")
    Observable<ArtistsCursorPager> getFollowedArtists(@Query("limit") int limit);

    @GET("/me/following?type=artist")
    Observable<ArtistsCursorPager> getFollowedArtists(@Query("limit") int limit, @Query("after") String spotifyId);

}

and

mSpotifyService.getFollowedArtists(50)
        .flatMap(result -> Observable.from(result.artists.items))
        .flatMap(this::responseToArtist)
        .sorted()
        .toList()
        .subscribe(new Subscriber<List<Artist>>() {
            @Override
            public void onNext(List<Artist> artists) {
                callback.onSuccess(artists);
            }
            // ...
        });

我想返回所有（在本例中为 119）艺术家callback.success(List<Artist>)。我是 RxJava 新手，所以我不确定是否有smart方法来做到这一点。

递归解决方案的唯一问题是堆栈溢出问题。一种无需递归的方法是

Observable<ArtistsCursorPager> allPages = Observable.defer(() ->
{
    BehaviorSubject<Object> pagecontrol = BehaviorSubject.create("start");
    Observable<ArtistsCursorPager> ret = pageControl.asObservable().concatMap(aKey ->
    {
        if (aKey != null && aKey.equals("start")) {
            return Observable.getFollowedArtists(50).doOnNext(page -> pagecontrol.onNext(page.cursors.after));
        } else if (aKey != null && !aKey.equals("")) {
            return Observable.getFollowedArtists(50,aKey).doOnNext(page -> pagecontrol.onNext(page.cursors.after));
        } else {
            return Observable.<ArtistsCursorPager>empty().doOnCompleted(()->pagecontrol.onCompleted());
        }        
    });
    return ret;
});

查看解决方案这个问题 https://stackoverflow.com/questions/37326380/paginate-observable-results-without-recursion-rxjava.