我有显示项目列表的活动,也有过滤和搜索选项。我正在使用 android 分页库显示项目。当我滚动到下一组正在加载的项目底部时,第一次加载项目列表时工作正常。但我也想过滤项目并搜索项目。在过滤或搜索项目时,我正在使现有源无效。如果不是使数据源无效,则过滤器和搜索 api 不会触发。我想使用数据源根据我的过滤器和搜索键加载新项目列表。
executor = Executors.newFixedThreadPool(5);
celebrityDataFactory = new CelebrityDataFactory(apicallInterface, mFansismParam);
networkState = Transformations.switchMap(celebrityDataFactory.getCelebrityData(),
dataSource -> dataSource.getNetworkState());
PagedList.Config pagedListConfig =
(new PagedList.Config.Builder())
.setEnablePlaceholders(false)
.setPrefetchDistance(8)
.setInitialLoadSizeHint(10)
.setPageSize(20).build();
if (!mFansismParam.getCategoryId().isEmpty()) {
celebrityDetails = new LivePagedListBuilder(celebrityDataFactory, pagedListConfig)
.setFetchExecutor(executor)
.build();
} else(!mFansismParam.getProfessionId().isEmpty()) {
celebrityDetails = new LivePagedListBuilder(celebrityDataFactory, pagedListConfig)
.setFetchExecutor(executor)
.build();
}
数据工厂创建数据源
@Override
public DataSource create() {
celebrityDataSource = new CelebrityDataSource(apicallInterface, params);
celebrityData.postValue(celebrityDataSource);
return celebrityDataSource;
}
改造 API 调用:
Call<CelebrityList> getCelebrityList(@Query("categoryId") String categoryId,
@Query("professionId") String professionId,
@Query("page") String pageNumber,
@Query("name") String searchKey);
数据源 API 回调:
apicallInterface.getCelebrityList(requestParams.getCategoryId(), "", "1", "").enqueue(new Callback<CelebrityList>() {
@Override
public void onResponse(Call<CelebrityList> call, Response<CelebrityList> response) {
if (response.isSuccessful()) {
initialLoading.postValue(NetworkState.LOADED);
networkState.postValue(NetworkState.LOADED);
if (!response.body().getData().isEmpty()) {
callback.onResult(response.body().getData(), null, "2");
} else {
networkState.postValue(new NetworkState(NetworkState.Status.SUCCESS, "No more results"));
}
} else {
initialLoading.postValue(new NetworkState(NetworkState.Status.FAILED, response.message()));
networkState.postValue(new NetworkState(NetworkState.Status.FAILED, response.message()));
}
}