我想进行多个API调用(有三个不同的查询)并合并结果,然后在onNext()
中显示它们。它有效,但我担心flatMap不适合这个。
@GET("www.examle.com/api/data/")
Observable<WebResultsResponse> getWebResults(@Query("param1") String query);
-----
private List<WebResult> resultsList;
private void requestWebResults(String query) {
resultsList.clear();
final Observable<List<WebResult>> observable = MainApplication.apiProvider.getApiProviderA.getWebResults("query1")
.subscribeOn(Schedulers.io())
.flatMap(new Function<WebResultsResponse, ObservableSource<List<WebResult>>>() {
@Override
public ObservableSource<List<WebResult>> apply(WebResultsResponse response) throws Exception {
if(response.getData() != null && response.getData().getResults() != null)
resultsList.addAll(response.getData().getResults());
return MainApplication.apiProvider.getApiProviderA.getWebResults("query2")
.flatMap(new Function<WebResultsResponse, ObservableSource<List<WebResult>>>() {
@Override
public ObservableSource<List<WebResult>> apply(WebResultsResponse response) throws Exception {
if(response.getData() != null && response.getData().getResults() != null)
resultsList.addAll(response.getData().getResults());
return MainApplication.apiProvider.getApiProviderA.getWebResults("query3")
.flatMap(new Function<WebResultsResponse, ObservableSource<List<WebResult>>>() {
@Override
public ObservableSource<List<WebResult>> apply(WebResultsResponse response) throws Exception {
if(response.getData() != null && response.getData().getResults() != null)
resultsList.addAll(response.getData().getResults());
return Observable.just(resultsList);
}
});
}
});
}
})
.observeOn(AndroidSchedulers.mainThread());
observer = new DisposableObserver<List<WebResult>>() {
@Override
public void onNext(List<WebResult> results) {
// do something with results
}
@Override
public void onError(Throwable e) {
}
@Override
public void onComplete() {
}
};
observable.subscribe(observer);
}
这是flatMap()
的正确使用吗?我可以以某种方式将resultsList
传递给链而不是将其声明为全局变量吗?
答案 0 :(得分:2)
如果您不关心哪一个首先返回,您可以简单地合并它们
Observable<List<WebResult>> observable = MainApplication.apiProvider.getApiProviderA.getWebResults("query1")
.mergeWith(MainApplication.apiProvider.getApiProviderA.getWebResults("query2"))
.mergeWith(MainApplication.apiProvider.getApiProviderA.getWebResults("query3"))
.subscribeOn(Schedulers.io())
.observeOn(AndroidSchedulers.mainThread());
在onNext()中你会得到每个结果作为它自己的列表,如果你想在完成所有可以使用的结果后得到结果
Observable<List<WebResult>> observable = Observable.zip(MainApplication.apiProvider.getApiProviderA.getWebResults("query1"), MainApplication.apiProvider.getApiProviderA.getWebResults("query2"), MainApplication.apiProvider.getApiProviderA.getWebResults("query3"), (webResults, webResults2, webResults3) -> {
List<WebResult> allResults = new ArrayList<>();
allResults.addAll(webResults);
allResults.addAll(webResults2);
allResults.addAll(webResults3);
return allResults;
});
在onNext()中,您将得到一个排放,所有结果加在一起
答案 1 :(得分:0)
在elmorabea's answer的帮助下,我使用zip
找到了这个解决方案:
List<Observable<WebResultsResponse>> results = new ArrayList<>();
results.add(MainApplication.apiProvider.getApiProviderA.getWebResults("query1"));
results.add(MainApplication.apiProvider.getApiProviderA.getWebResults("query2"));
results.add(MainApplication.apiProvider.getApiProviderA.getWebResults("query3"));
Observable<List<WebResult>> observable = Observable.zip(results, new Function<Object[], List<WebResult>>() {
@Override
public List<WebResult> apply(Object[] responses) throws Exception {
List<WebResult> allResults = new ArrayList<>();
for(int i=0; i<responses.length; i++) {
WebResultsResponse response = (WebResultsResponse)responses[i];
if(response != null && response.getData() != null && response.getData().getResults() != null)
allResults.addAll(response.getData().getResults());
}
return allResults;
}
})
.observeOn(AndroidSchedulers.mainThread())
.subscribeOn(Schedulers.io());