Question

我是rx-java和rx-android的完全初学者。我听说学习曲线在开始时非常陡峭。

我试图通过使用rx-android替换所有基于Eventbus的代码到更安全的替代方案。

我已设置此代码段以根据编辑文本文本更改事件创建可观察对象：

MainActivity

814788aa9b6a22bf73ff3ae5fe0c6e0c, 0.15059251440334190624
814788aa9b6a22bf73ff3ae5fe0c6e0c, 0.599991561440334190624
814788aa9b6a22bf73ff3ae5fe0c6e0c, 0.54911111440334190624
814788aa9b6a22bf73ff3ae5fe0c6e0c, 0.0982222561440334190624
814788aa9b6a22bf73ff3ae5fe0c6e0c, 0.169987141440334190624
814788aa9b6a22bf73ff3ae5fe0c6e0c, 0.032851281440334190624
814788aa9b6a22bf73ff3ae5fe0c6e0c, 0.223390041440334190624
814788aa9b6a22bf73ff3ae5fe0c6e0c, 0.70173611440334190624
....
....
....

RxUtils：

RxUtils.createEditTextChangeObservable(txtInput).throttleLast(200, TimeUnit.MILLISECONDS, AndroidSchedulers.mainThread()).subscribe(new Action1<EditText>() {
            @Override
            public void call(EditText editText) {
                searchStopResultFragment.query(editText.getText().toString());
            }
        });

SearchStopResultFragment：

public static Observable<EditText> createEditTextChangeObservable(final EditText editText){
        return Observable.create(new Observable.OnSubscribe<EditText>() {
            @Override
            public void call(final Subscriber<? super EditText> subscriber) {
                editText.addTextChangedListener(new TextWatcher() {
                    @Override
                    public void beforeTextChanged(CharSequence s, int start, int count, int after) {

                    }

                    @Override
                    public void onTextChanged(CharSequence s, int start, int before, int count) {

                    }

                    @Override
                    public void afterTextChanged(Editable s) {
                        if (subscriber.isUnsubscribed()) return;
                        subscriber.onNext(editText);
                    }
                });
            }
        });
    }

感觉我做错了。我在每个文本更改事件的片段中从查询方法创建新的observable。我还想根据public void query(String query){ lastQuery = query; resultObservable = StopProvider.getStopResultObservable(getActivity().getContentResolver(),query); subscription = resultObservable.subscribeOn(Schedulers.newThread()) .observeOn(AndroidSchedulers.mainThread()).subscribe(new Observer<List<Stop>>() { @Override public void onCompleted() { } @Override public void onError(Throwable e) { } @Override public void onNext(List<Stop> stops) { if(!lastQuery.equals("")) { if(stops.size()>0) { ArrayList<AdapterItem> items = adapter.getItems(); items.clear(); for (Stop stop : stops) { SearchResultStopItem item = new SearchResultStopItem(stop, SearchResultStopItem.STOP); items.add(item); } adapter.setItems(items); adapter.notifyDataSetChanged(); }else{ //DO A NOTHER ASYNC QUERY TO FETCH RESULTS } }else{ showStartItems(); } } }); }中的结果创建一个新的异步查找操作（参见注释）

有没有？

Answer 1

您可以将第二个concatMap移动到您需要的唯一位置 - 在combineLatest

之后

    RxUtils.createEditTextChangeObservable(txtInput)
            .throttleLast(200, TimeUnit.MILLISECONDS, AndroidSchedulers.mainThread())
            .concatMap(new Func1<EditText, Observable<Pair<String, List<Stop>>>>() {
                @Override
                public Observable<Pair<String, List<Stop>>> call(EditText editText) {
                    String query = editText.getText().toString();
                    //searchStopResultFragment.setLastQuery(query);
                    if (query.isEmpty()) {
                        return Observable.just(null);
                    }
                    return Observable
                            .combineLatest(StopProvider.getStopResultObservable(getContentResolver(), query), Observable.just(query), new Func2<List<Stop>, String, Pair<String, List<Stop>>>() {
                                @Override
                                public Pair<String, List<Stop>> call(List<Stop> stops, String s) {
                                    return new Pair(s, stops);
                                }
                            })
                            .concatMap(new Func1<R, Observable<? extends Pair<String, List<Stop>>>>() {
                                @Override
                                public Observable<? extends Pair<String, List<Stop>>> call(R r) {
                                    if (queryAndStops.second.size() == 0) {
                                        return RestClient.service().locationName(queryAndStops.first).concatMap(new Func1<LocationNameResponse, Observable<? extends List<Stop>>>() {
                                            @Override
                                            public Observable<? extends List<Stop>> call(LocationNameResponse locationNameResponse) {
                                                return Observable.just(locationNameResponse.getAddresses());
                                            }
                                        });
                                    } else {
                                        return Observable.just(queryAndStops.second);
                                    }
                                }
                            });
                }
            })
            .subscribeOn(Schedulers.newThread())
            .observeOn(AndroidSchedulers.mainThread()).compose(this.<List<Stop>>bindToLifecycle())
            .subscribe(new Action1<List<Stop>>() {
                @Override
                public void call(List<Stop> stops) {
                    if (stops != null) {
                        searchStopResultFragment.showStops(stops);
                    } else {
                        searchStopResultFragment.showStartItems();
                    }

                }
            }, new Action1<Throwable>() {
                @Override
                public void call(Throwable throwable) {
                    showError(throwable);
                }
            });

Answer 2

以下是我提出的建议：

RxUtils.createEditTextChangeObservable(txtInput)
.throttleLast(200, TimeUnit.MILLISECONDS, AndroidSchedulers.mainThread())
.map(EXTRACT_STRING)
.filter(STRING_IS_NOT_EMPTY)
.concatMap(new Func1<EditText, Observable<Pair<String,List<Stop>>>>() {

    @Override
    public Observable<Pair<String, List<Stop>>> call(final String query) {

        return StopProvider.getStopResultObservable(getContentResolver(), query)
        .map(new Func1<List<Stop>, Pair<String, List<Stop>>>() {
            // I think this map is a bit more readable than the 
            // combineLatest, and since "query" should not be changing 
            // anyway, the result should be the same (you have to 
            // declare it as final in the method signature, though
            @Override
            public Pair<String, List<Stop>> call(List<Stop> stops) {
                return new Pair(query, stops);
            }
        });
    }
)
.concatMap(new Func1<Pair<String, List<Stop>>, Observable<List<Stop>>>() {

    @Override
    public Observable<List<Stop>> call(Pair<String, List<Stop>> queryAndStops) {
        if (queryAndStops.second.size() == 0) {
            return RestClient.service().locationName(queryAndStops.first)
                       .map(new Func1<LocationNameResponse, List<Stop>>() {

                            @Override
                            public List<Stop> call(LocationNameResponse locationNameResponse) {
                                // since there was no if-else in your original code (you were always
                                // just wrapping the List in an Observable) I removed that, too
                                return locationNameResponse.getAddresses();
                            }
            });
        } else {
            return Observable.just(queryAndStops.second);
        }
    }
)
.subscribeOn(Schedulers.newThread())
.observeOn(AndroidSchedulers.mainThread())
.compose(this.<List<Stop>>bindToLifecycle())
.subscribe(new Action1<List<Stop>>() {
    @Override
    public void call(List<Stop> stops) {
        // since I don't know what your API is returning I think
        // it's saver to keep this check in:
        if (stops != null) {
            searchStopResultFragment.showStops(stops);
        } else {
            searchStopResultFragment.showStartItems();
        }
    }
},
new Action1<Throwable>() {
    @Override
    public void call(Throwable throwable) {
        showError(throwable);
    }
});

其中：

public static final Func1<EditText, String> EXTRACT_STRING = new Func1<EditText, String>() {

    @Override
    public void String call(EditText editText) {
        return editText.getText().toString();
    }
};

public static final Func1<String, Boolean> STRING_IS_NOT_EMPTY = new Func1<String, Boolean>() {

    @Override
    public void String call(String string) {
        return !string.isEmpty();
    }
};

所以，这至少不需要返回Observable.just(null)，然后检查链中的那个。

根据结果链接可观察量

2 个答案:

根据结果​​链接可观察量

2 个答案:

根据结果链接可观察量