Android 在RxJava中定义订户顺序的方法?
我正在寻找一种定义观察者顺序(?)的方法Android 在RxJava中定义订户顺序的方法?,android,retrofit,rx-java,Android,Retrofit,Rx Java,我正在寻找一种定义观察者顺序(?)的方法 @GET("/get_user_msgs") Observable<PrivateMessagesResponse> getPrivateMessages(@QueryMap Map<String, String> params); 我还有一个用于我的浓缩咖啡测试的API包装,我在那里订阅了相同的Observable。这样,API包装器中的观察者将首先被调用,然后才是ListView中的观察者 被称为 public class
@GET("/get_user_msgs")
Observable<PrivateMessagesResponse> getPrivateMessages(@QueryMap Map<String, String> params);
我还有一个用于我的浓缩咖啡测试的API包装,我在那里订阅了相同的Observable
。这样,API包装器中的观察者将首先被调用,然后才是ListView中的观察者
被称为
public class IdlingWrapper implements Api, IdlingResource {
....
public IdlingWrapper(Api realApi) {
this.realApi = realApi;
}
...
public Observable<PrivateMessagesResponse> getPrivateMessages(@QueryMap Map<String, String> params); {
counter.incrementAndGet();
return wrapObservable(realApi.getPrivateMessages(params));
}
protected <T> Observable<T> wrapObservable(final Observable<PrivateMessagesResponse> observable) {
//what to do here?
}
}
因此,observer a
将首先得到通知,然后是observer b
,然后才是LastObserver
或者其他任何我可以知道所有注册观察员何时得到通知并完成的方法。看来,这样做效果不错
protected <T> Observable<T> wrapObservable(final Observable<T> original) {
return Observable.create(new Observable.OnSubscribeFunc<T>() {
@Override
public Subscription onSubscribe(final Observer<? super T> t1) {
original.subscribe(new Observer<T>() {
@Override
public void onCompleted() {
t1.onCompleted();
uiThreadHandler.post(new Runnable() {
@Override
public void run() {
counter.decrementAndGet();
notifyIdle();
}
});
}
@Override
public void onError(Throwable e) {
t1.onError(e);
uiThreadHandler.post(new Runnable() {
@Override
public void run() {
counter.decrementAndGet();
notifyIdle();
}
});
}
@Override
public void onNext(T args) {
t1.onNext(args);
}
});
return Subscriptions.empty();
}
});
}
受保护的可观察包装可观察(最终可观察原件){
return Observable.create(new Observable.onSubscribeeFunc()){
@凌驾
publicsubscriptiononsubscribe(finalobserver我不确定您在IdlingWrapper中想要做什么,但我认为当前的实现非常脆弱
我认为最重要的事情是要保证可观测数据只能被调用一次
下面是一个快速实现来演示这一点,以及我对wrapObservable的实现
public class Test {
private static int counter = 0;
private static final List<Observable<?>> list = Collections.synchronizedList(new ArrayList<>());
protected static <T> Observable<T> wrapObservable(final Observable<T> original) {
// run atleast once???
synchronized (list) {
list.add(original);
}
return Observable.create(new Observable.OnSubscribe<Void>() {
@Override
public void call(Subscriber<? super Void> subscriber) {
synchronized (list) {
counter++;
if (!list.contains(original)) {
subscriber.onError(new Exception("You can only subscribe once!"));
return;
}
list.remove(original);
}
// Sleep to make it easier to see things happening...
try {
Thread.sleep(3000);
} catch (InterruptedException ignored) {
}
subscriber.onCompleted();
}
}).flatMap(new Func1<Void, Observable<? extends T>>() {
@Override
public Observable<? extends T> call(Void o) {
return original;
}
}).finallyDo(new Action0() {
@Override
public void call() {
synchronized (list) {
counter--;
if (list.size() == 0 && counter == 0) {
System.err.println("finally");
}
}
}
});
}
public static void main(String[] args) throws InterruptedException {
for(int i = 0; i < 10; i++) {
// running in io thread for simulating async call.
Observable<String> test = wrapObservable(Observable.from("TEST!!!!!!")).subscribeOn(Schedulers.io());
test.subscribe(new Observer<String>() {
@Override
public void onCompleted() {
System.err.println("completed");
}
@Override
public void onError(Throwable e) {
System.err.println("error");
}
@Override
public void onNext(String s) {
System.err.println("next");
}
});
// example of calling the same observable twice.
test.subscribe(new Observer<String>() {
@Override
public void onCompleted() {
System.err.println("completed");
}
@Override
public void onError(Throwable e) {
System.err.println("error");
}
@Override
public void onNext(String s) {
System.err.println("next");
}
});
}
Thread.sleep(10000);
}
}
公共类测试{
专用静态整数计数器=0;
私有静态最终列表如果您只想使用内置的RxJava方法对观察者进行排序,您可以使用flatMap和range将每个项目转换为多个具有优先级的项目,然后根据优先级进行筛选。观察者的排序基于其筛选方式
下面是一个简单的例子:
Observable<Pair<Integer, Object>> shared = RxView.clicks(findViewById(R.id.textView))
.flatMap(c -> Observable.range(0, 2).map(i -> Pair.create(i, c)))
.share();
shared.filter(p -> p.first == 1)
.map(p -> p.second)
.doOnSubscribe(c -> Log.d(TAG, "first subscribed doOnSubscribe"))
.subscribe(c -> Log.d(TAG, "first subscribed onNext"));
shared.filter(p -> p.first == 0)
.map(p -> p.second)
.doOnSubscribe(c -> Log.d(TAG, "second subscribed doOnSubscribe"))
.subscribe(c -> Log.d(TAG, "second subscribed onNext"));
Observable shared=RxView.clicks(findviewbyd(R.id.textView))
.flatMap(c->Observable.range(0,2).map(i->Pair.create(i,c)))
.share();
shared.filter(p->p.first==1)
.map(p->p.second)
.doOnSubscribe(c->Log.d(标记“首次订阅的doOnSubscribe”))
.subscribe(c->Log.d(标记“first subscribed onNext”);
shared.filter(p->p.first==0)
.map(p->p.second)
.doOnSubscribe(c->Log.d(标记“第二次订阅的doOnSubscribe”))
.subscribe(c->Log.d(标记为“第二个subscribed onNext”);
如果您在各地都这样做我想您可以创建一个类似于“OperatorDoonech”的新操作符:这是怎么回事?我没有访问observer a
或observer b
的权限,所以你想调用observer a,observer b,然后是LastObserver?看起来这是不可能的。一个可观察者怎么知道没有其他观察者?我真的不需要LastObserver
我只想知道所有注册的观察者何时开始(在这种情况下,observer a
和observer b
)已通知并完成或处理了错误。我将更新问题。查看最终结果。包装器正在执行该方法的目的。我正在尝试确保,IdlingWrapper
中的我的观察者将被最后调用,而不是订阅原始观察者的任何其他观察者。是的,如果存在,我的解决方案将中断在原始的observable上有2个订户。这将更好地工作。谢谢!但在那个同步列表上不是一个粉丝。。。
public class Test {
private static int counter = 0;
private static final List<Observable<?>> list = Collections.synchronizedList(new ArrayList<>());
protected static <T> Observable<T> wrapObservable(final Observable<T> original) {
// run atleast once???
synchronized (list) {
list.add(original);
}
return Observable.create(new Observable.OnSubscribe<Void>() {
@Override
public void call(Subscriber<? super Void> subscriber) {
synchronized (list) {
counter++;
if (!list.contains(original)) {
subscriber.onError(new Exception("You can only subscribe once!"));
return;
}
list.remove(original);
}
// Sleep to make it easier to see things happening...
try {
Thread.sleep(3000);
} catch (InterruptedException ignored) {
}
subscriber.onCompleted();
}
}).flatMap(new Func1<Void, Observable<? extends T>>() {
@Override
public Observable<? extends T> call(Void o) {
return original;
}
}).finallyDo(new Action0() {
@Override
public void call() {
synchronized (list) {
counter--;
if (list.size() == 0 && counter == 0) {
System.err.println("finally");
}
}
}
});
}
public static void main(String[] args) throws InterruptedException {
for(int i = 0; i < 10; i++) {
// running in io thread for simulating async call.
Observable<String> test = wrapObservable(Observable.from("TEST!!!!!!")).subscribeOn(Schedulers.io());
test.subscribe(new Observer<String>() {
@Override
public void onCompleted() {
System.err.println("completed");
}
@Override
public void onError(Throwable e) {
System.err.println("error");
}
@Override
public void onNext(String s) {
System.err.println("next");
}
});
// example of calling the same observable twice.
test.subscribe(new Observer<String>() {
@Override
public void onCompleted() {
System.err.println("completed");
}
@Override
public void onError(Throwable e) {
System.err.println("error");
}
@Override
public void onNext(String s) {
System.err.println("next");
}
});
}
Thread.sleep(10000);
}
}
Observable<Pair<Integer, Object>> shared = RxView.clicks(findViewById(R.id.textView))
.flatMap(c -> Observable.range(0, 2).map(i -> Pair.create(i, c)))
.share();
shared.filter(p -> p.first == 1)
.map(p -> p.second)
.doOnSubscribe(c -> Log.d(TAG, "first subscribed doOnSubscribe"))
.subscribe(c -> Log.d(TAG, "first subscribed onNext"));
shared.filter(p -> p.first == 0)
.map(p -> p.second)
.doOnSubscribe(c -> Log.d(TAG, "second subscribed doOnSubscribe"))
.subscribe(c -> Log.d(TAG, "second subscribed onNext"));