Android RxJava加入列表
试图理解所有RxJava的东西。我在做下面的例子:Android RxJava加入列表,android,rx-java,Android,Rx Java,试图理解所有RxJava的东西。我在做下面的例子: private Observable<List<String>> query1() { List<String> urls = new ArrayList<>(); urls.add("1"); urls.add("2"); urls.add("3"); urls.add("4"); return Observable.just(urls); }
private Observable<List<String>> query1() {
List<String> urls = new ArrayList<>();
urls.add("1");
urls.add("2");
urls.add("3");
urls.add("4");
return Observable.just(urls);
}
private Observable<List<String>> query2() {
List<String> urls = new ArrayList<>();
urls.add("A");
urls.add("B");
urls.add("C");
urls.add("D");
return Observable.just(urls);
}
私有可观察查询1(){
列表URL=新的ArrayList();
添加(“1”);
添加(“2”);
添加(“3”);
添加(“4”);
返回可观察的.just(URL);
}
私有可观察查询2(){
列表URL=新的ArrayList();
URL。添加(“A”);
URL。添加(“B”);
添加(“C”);
URL。添加(“D”);
返回可观察的.just(URL);
}
然后尝试加入两个列表:
Observable.zip(
query1(),
query2(),
new Func2<List<String>, List<String>, Observable<String>>() {
@Override
public Observable<String> call(List<String> a1, List<String> a2) {
List<String> list = new ArrayList<>();
list.addAll(a1);
list.addAll(a2);
return Observable.from(list);
}
})
.subscribe(new Action1<String>() { // <-- It says, cannot resolve method subscribe
@Override
public void call(String string) {
String text = testTextView.getText().toString();
testTextView.setText(text + "\n" + string);
}
});
Observable.zip(
查询1(),
query2(),
新功能2(){
@凌驾
公共可观测呼叫(列表a1、列表a2){
列表=新的ArrayList();
列表。添加全部(a1);
列表。addAll(a2);
从(列表)中可观察到的返回;
}
})
.subscribe(new Action1(){/这是因为您试图从zip函数返回Observable,但随后您传递了Action
Observable.zip(
查询1(),
query2(),
新功能2(){
@凌驾
公共列表呼叫(列表a1、列表a2){
列表=新的ArrayList();
列表。添加全部(a1);
列表。addAll(a2);
退货清单;
}
})
.订阅(
(字符串)->System.out.println(字符串)
);
我相信您正在寻找的运营商是concat
或merge
Concat
将从两个或多个可观察的
s发出排放,而不交错它们
另一方面,Merge
将通过合并其发射来合并多个观测值
例如:
String[] numbers = {"1", "2", "3", "4"};
String[] letters = {"a", "b", "c", "d"};
Observable<String> query1 = Observable.from(numbers).delay(1, TimeUnit.SECONDS);
Observable<String> query2 = Observable.from(letters);
Observable
.concat(query1, query2)
.subscribe(s -> {
System.out.printf("-%s-" + s);
});
将输出-(1,a)-(2,b)-(3,c)-(4,d)
可观察查询1(){
列表s=新的ArrayList();
s、 添加(“1”);s.add(“1”);s.add(“1”);
可观察到的回报。仅(s);
}
可观测查询2(){
列表s=新的ArrayList();
s、 添加(“1”);s.add(“1”);s.add(“1”);
可观察到的回报。仅(s);
}
void HelloRx(){
Map map2=new LinkedHashMap();//在此处选择要返回的结果!
zip(query1(),//可观察方法1
query2(),//可观察方法2
(结果1,结果2)->{
for(String s:result1){//result1是由query1、result2…返回的值。
//你想干什么就干什么
//地图放置(……)
}
返回null;
})
.subscribeOn(BackgroundSchedulers.getMultiThreadInstance())
.observeOn(AndroidSchedulers.mainThread())
.doOnCompleted(()->{
//完成后执行一些操作,例如将数据传输到适配器
})
.subscribe();
}
显然,要将两个列表合并为一个列表,您可以对它们的Observable.from()
执行Observable.concat()
,然后调用Observable.toList()
RealmResults equalTo;
真正的结果始于;
@凌驾
公共void onViewRestored(){
compositeSubscription=新compositeSubscription();
equalTo=realm.where(Cat.class).equalTo(“字段”,filterString.findAllSorted(“字段”);
beginsWith=realm.where(Cat.class).beginsWith(“field”,filterString.findAllSorted(“field”);
compositeSubscription.add(realm.asObservable()
.switchMap(新函数1(){
@凌驾
公共可观察调用(领域){
返回可观测的concat(可观测的from(equalTo),可观测的from(beginsWith));
}
})
托利斯先生()
.订阅(猫->{
//使用列表更新适配器
}));
我不得不将flatMap添加到您发布的内容中(编辑了我的问题)。感谢concat
示例。
Observable.zip(
query1(),
query2(),
new Func2<List<String>, List<String>, List<String>>() {
@Override
public List<String> call(List<String> a1, List<String> a2) {
List<String> list = new ArrayList<>();
list.addAll(a1);
list.addAll(a2);
return list;
}
})
.subscribe(
(string)-> System.out.println(string)
);
String[] numbers = {"1", "2", "3", "4"};
String[] letters = {"a", "b", "c", "d"};
Observable<String> query1 = Observable.from(numbers).delay(1, TimeUnit.SECONDS);
Observable<String> query2 = Observable.from(letters);
Observable
.concat(query1, query2)
.subscribe(s -> {
System.out.printf("-%s-" + s);
});
Observable
.zip(query1, query2, (String n, String l) -> String.format("(%s, %s)", n, l))
.subscribe(s -> {
System.out.printf("-%s-", s);
});
Observable<List<String>> query1(){
List<String> s = new ArrayList<>();
s.add("1");s.add("1");s.add("1");
return Observable.just(s);
}
Observable<List<String>> query2(){
List<String> s = new ArrayList<>();
s.add("1");s.add("1");s.add("1");
return Observable.just(s);
}
void HelloRx(){
Map<String,String> map2=new LinkedHashMap<>();//pick the result you want to return Here !
Observable.zip(query1(),//Observable Method 1
query2(),//Observable Method 2
(result1,result2)->{
for(String s : result1){//result1 is the value returned by query1 , result2 ...u know.
//do whatever you want
//map.put(......)
}
return null;
})
.subscribeOn(BackgroundSchedulers.getMultiThreadInstance())
.observeOn(AndroidSchedulers.mainThread())
.doOnCompleted(() -> {
//Do Something when finish for example transmit data to your adapter
})
.subscribe();
}
RealmResults<Cat> equalTo;
RealmResults<Cat> beginsWith;
@Override
public void onViewRestored() {
compositeSubscription = new CompositeSubscription();
equalTo = realm.where(Cat.class).equalTo("field", filterString).findAllSorted("field");
beginsWith = realm.where(Cat.class).beginsWith("field", filterString).findAllSorted("field");
compositeSubscription.add(realm.asObservable()
.switchMap(new Func1<Realm, Observable<Cat>>() {
@Override
public Observable<Cat> call(Realm realm) {
return Observable.concat(Observable.from(equalTo), Observable.from(beginsWith));
}
})
.toList()
.subscribe(cats -> {
// update adapter with List<Cat>
}));