为什么Single.concat(..,..)。toObservable()不只返回一个可观察到的观测值?
我正在从不同来源以相同结果类型访问Android Room数据库中的搜索功能。
我想按结果顺序显示第一个来源,然后第二个来源。
@Query("SELECT id, title, from table1 where title LIKE :title || '%'")
fun getItemsFirstTable(title: String): Single<List<MinimumResult>>
@Query("SELECT id, title, from table2 where title LIKE :title || '%'")
fun getItemsSecondTable(title: String): Single<List<MinimumResult>>
用于合并Single并在视图模型中返回Observable的方法
public Observable<List<SearchItem>> concatSearchedItems(String s){
return Single.concat(UseCaseKt.getItemsFirstTable(s), UseCaseKt.getItemsSecondTable(s)).toObservable()
}
认同TextWatcher
Observable.create(new ObservableOnSubscribe<Object>() {
@Override
public void subscribe(ObservableEmitter<Object> emitter) throws Exception {
mEditTextSearch.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) {
emitter.onNext(s.toString());
}
@Override
public void afterTextChanged(Editable s) {
}
});
}
})
.subscribeOn(Schedulers.io())
.doOnNext( l -> Log.d(TAG, "onCreateView: "))
.debounce(500, TimeUnit.MILLISECONDS)
.distinctUntilChanged()
.switchMap(new Function<Object, ObservableSource<List<SearchItem>>>() {
@Override
public ObservableSource<List<SearchItem>> apply(Object o) throws Exception {
return searchViewModel.concatSearchedItems(o.toString());
}
})
.observeOn(AndroidSchedulers.mainThread())
.subscribe(new Observer<List<SearchItem>>() {
@Override
public void onSubscribe(Disposable d) {
}
@Override
public void onNext(List<SearchItem> searchItems) {
searchViewModel.searchList.setValue(searchItems);
}
@Override
public void onError(Throwable e) {
}
@Override
public void onComplete() {
}
});
调试之后,我发现onNext被调用了两次,尽管我已经连接了单个结果。
答案 0 :(得分:1)
public Observable<List<SearchItem>> concatSearchedItems(String s){
return Singles.zip(UseCaseKt.getItemsFirstTable(s), UseCaseKt.getItemsSecondTable(s))
{ tab1list, tab2list -> listOf<SearchItem>().plus(tab1list).plus(tab2list) }
.toObservable()
}
不会从Singles发出的数据中合并数据,而是concat自身进行的Singles。
因此,您需要使用zip:
<a
class="link"
href="https://google.com/same/index.html#ancor"
target="_blank">
{{ $t("labelLearnMore") }}
</a>