Zip操作符会严格地将可观察对象发出的项配对。它会等待两个（或更多）项到达，然后将它们合并。因此，这对您的需求是合适的。
我会使用Func2将前两个可观察对象的结果链接起来。
请注意，如果使用Retrofit，此方法会更简单，因为其API接口可能会返回一个可观察对象。否则，您需要创建自己的可观察对象。
// assuming each observable returns response in the form of String
Observable<String> movOb = Observable.create(...);
// if you use Retrofit
Observable<String> picOb = RetrofitApiManager.getService().uploadPic(...),
Observable.zip(movOb, picOb, new Func2<String, String, MyResult>() {
@Override
public MyResult call(String movieUploadResponse, String picUploadResponse) {
// analyze both responses, upload them to another server
// and return this method with a MyResult type
return myResult;
}
}
)
// continue chaining this observable with subscriber
// or use it for something else
一个小例子：
val observableOne = Observable.just("Hello", "World")
val observableTwo = Observable.just("Bye", "Friends")
val zipper = BiFunction<String, String, String> { first, second -> "$first - $second" }
Observable.zip(observableOne, observableTwo, zipper)
.subscribe { println(it) }
这将打印：
Hello - Bye
World - Friends
BiFunction<String, String, String>表示第一个String是第一个observable的类型，第二个String是第二个observable的类型，第三个String代表你的zipper函数返回的类型。
我在这篇博客文章中制作了一个小例子，使用zip调用了两个真实的端点。
Func2<...> 中，这三个字符串是什么？是为了 s、- 和 s2 吗？ - chornge我这里有一个使用异步方式完成的Zip例子，以防你感到好奇
/**
* Since every observable into the zip is created to subscribeOn a diferent thread, it´s means all of them will run in parallel.
* By default Rx is not async, only if you explicitly use subscribeOn.
*/
@Test
public void testAsyncZip() {
scheduler = Schedulers.newThread();
scheduler1 = Schedulers.newThread();
scheduler2 = Schedulers.newThread();
long start = System.currentTimeMillis();
Observable.zip(obAsyncString(), obAsyncString1(), obAsyncString2(), (s, s2, s3) -> s.concat(s2)
.concat(s3))
.subscribe(result -> showResult("Async in:", start, result));
}
/**
* In this example the the three observables will be emitted sequentially and the three items will be passed to the pipeline
*/
@Test
public void testZip() {
long start = System.currentTimeMillis();
Observable.zip(obString(), obString1(), obString2(), (s, s2, s3) -> s.concat(s2)
.concat(s3))
.subscribe(result -> showResult("Sync in:", start, result));
}
public void showResult(String transactionType, long start, String result) {
System.out.println(result + " " +
transactionType + String.valueOf(System.currentTimeMillis() - start));
}
public Observable<String> obString() {
return Observable.just("")
.doOnNext(val -> {
System.out.println("Thread " + Thread.currentThread()
.getName());
})
.map(val -> "Hello");
}
public Observable<String> obString1() {
return Observable.just("")
.doOnNext(val -> {
System.out.println("Thread " + Thread.currentThread()
.getName());
})
.map(val -> " World");
}
public Observable<String> obString2() {
return Observable.just("")
.doOnNext(val -> {
System.out.println("Thread " + Thread.currentThread()
.getName());
})
.map(val -> "!");
}
public Observable<String> obAsyncString() {
return Observable.just("")
.observeOn(scheduler)
.doOnNext(val -> {
System.out.println("Thread " + Thread.currentThread()
.getName());
})
.map(val -> "Hello");
}
public Observable<String> obAsyncString1() {
return Observable.just("")
.observeOn(scheduler1)
.doOnNext(val -> {
System.out.println("Thread " + Thread.currentThread()
.getName());
})
.map(val -> " World");
}
public Observable<String> obAsyncString2() {
return Observable.just("")
.observeOn(scheduler2)
.doOnNext(val -> {
System.out.println("Thread " + Thread.currentThread()
.getName());
})
.map(val -> "!");
}
zip 操作符允许您从两个不同的可观察对象的结果中组合一个结果。
您需要提供一个 lambda 函数，该函数将根据每个可观察对象发出的数据创建一个结果。
Observable<MovieResponse> movies = ...
Observable<PictureResponse> picture = ...
Observable<Response> response = movies.zipWith(picture, (movie, pic) -> {
return new Response("description", movie.getName(), pic.getUrl());
});
这是我的实现，使用了 Single.zip 和 rxJava2
我尽可能地让它易于理解
//
// API Client Interface
//
@GET(ServicesConstants.API_PREFIX + "questions/{id}/")
Single<Response<ResponseGeneric<List<ResponseQuestion>>>> getBaseQuestions(@Path("id") int personId);
@GET(ServicesConstants.API_PREFIX + "physician/{id}/")
Single<Response<ResponseGeneric<List<ResponsePhysician>>>> getPhysicianInfo(@Path("id") int personId);
//
// API middle layer - NOTE: I had feedback that the Single.create is not needed (but I haven't yet spent the time to improve it)
//
public Single<List<ResponsePhysician>> getPhysicianInfo(int personId) {
return Single.create(subscriber -> {
apiClient.getPhysicianInfo(appId)
.subscribeOn(Schedulers.io())
.observeOn(Schedulers.io())
.subscribe(response -> {
ResponseGeneric<List<ResponsePhysician>> responseBody = response.body();
if(responseBody != null && responseBody.statusCode == 1) {
if (!subscriber.isDisposed()) subscriber.onSuccess(responseBody.data);
} else if(response.body() != null && response.body().status != null ){
if (!subscriber.isDisposed()) subscriber.onError(new Throwable(response.body().status));
} else {
if (!subscriber.isDisposed()) subscriber.onError(new Throwable(response.message()));
}
}, throwable -> {
throwable.printStackTrace();
if(!subscriber.isDisposed()) subscriber.onError(throwable);
});
});
}
public Single<List<ResponseQuestion>> getHealthQuestions(int personId){
return Single.create(subscriber -> {
apiClient.getBaseQuestions(personId)
.subscribeOn(Schedulers.io())
.observeOn(Schedulers.io())
.subscribe(response -> {
ResponseGeneric<List<ResponseQuestion>> responseBody = response.body();
if(responseBody != null && responseBody.data != null) {
if (!subscriber.isDisposed()) subscriber.onSuccess(response.body().data);
} else if(response.body() != null && response.body().status != null ){
if (!subscriber.isDisposed()) subscriber.onError(new Throwable(response.body().status));
} else {
if (!subscriber.isDisposed()) subscriber.onError(new Throwable(response.message()));
}
}, throwable -> {
throwable.printStackTrace();
if(!subscriber.isDisposed()) subscriber.onError(throwable);
});
});
}
//please note that ResponseGeneric is just an outer wrapper of the returned data - common to all API's in this project
public class ResponseGeneric<T> {
@SerializedName("Status")
public String status;
@SerializedName("StatusCode")
public float statusCode;
@SerializedName("Data")
public T data;
}
//
// API end-use layer - this gets close to the UI so notice the oberver is set for main thread
//
private static class MergedResponse{// this is just a POJO to store all the responses in one object
public List<ResponseQuestion> listQuestions;
public List<ResponsePhysician> listPhysicians;
public MergedResponse(List<ResponseQuestion> listQuestions, List<ResponsePhysician> listPhysicians){
this.listQuestions = listQuestions;
this.listPhysicians = listPhysicians;
}
}
// example of Single.zip() - calls getHealthQuestions() and getPhysicianInfo() from API Middle Layer
private void downloadHealthQuestions(int personId) {
addRxSubscription(Single
.zip(getHealthQuestions(personId), getPhysicianInfo(personId), MergedResponse::new)
.subscribeOn(Schedulers.io())
.observeOn(AndroidSchedulers.mainThread())
.subscribe(response -> {
if(response != null) {
Timber.i(" - total health questions downloaded %d", response.listQuestions.size());
Timber.i(" - physicians downloaded %d", response.listPhysicians.size());
if (response.listPhysicians != null && response.listPhysicians.size()>0) {
// do your stuff to process response data
}
if (response.listQuestions != null && response.listQuestions.size()>0) {
// do your stuff to process response data
}
} else {
// process error - show message
}
}, error -> {
// process error - show network error message
}));
}
我一直在寻找一个简单的答案，关于如何使用Zip操作符以及如何处理创建的Observables以将它们传递给Zip操作符。我想知道是否应该为每个Observable调用subscribe()方法。但是没有一个简单的答案可以找到，我不得不自己摸索。下面是一个使用Zip操作符处理两个Observables的简单示例：
@Test
public void zipOperator() throws Exception {
List<Integer> indexes = Arrays.asList(0, 1, 2, 3, 4);
List<String> letters = Arrays.asList("a", "b", "c", "d", "e");
Observable<Integer> indexesObservable = Observable.fromIterable(indexes);
Observable<String> lettersObservable = Observable.fromIterable(letters);
Observable.zip(indexesObservable, lettersObservable, mergeEmittedItems())
.subscribe(printMergedItems());
}
@NonNull
private BiFunction<Integer, String, String> mergeEmittedItems() {
return new BiFunction<Integer, String, String>() {
@Override
public String apply(Integer index, String letter) throws Exception {
return "[" + index + "] " + letter;
}
};
}
@NonNull
private Consumer<String> printMergedItems() {
return new Consumer<String>() {
@Override
public void accept(String s) throws Exception {
System.out.println(s);
}
};
}
[0] a
[1] b
[2] c
[3] d
[4] e
你使用 Java 8 中的 rxjava 的 zip：
Observable<MovieResponse> movies = ...
Observable<PictureResponse> picture = ...
Observable<ZipResponse> response = Observable.zip(movies, picture, ZipResponse::new);
class ZipResponse {
private MovieResponse movieResponse;
private PictureResponse pictureResponse;
ZipResponse(MovieResponse movieResponse, PictureResponse pictureResponse) {
this.movieResponse = movieResponse;
this.pictureResponse = pictureResponse;
}
public MovieResponse getMovieResponse() {
return movieResponse;
}
public void setMovieResponse(MovieResponse movieResponse) {
this.movieResponse= movieResponse;
}
public PictureResponse getPictureResponse() {
return pictureResponse;
}
public void setPictureResponse(PictureResponse pictureResponse) {
this.pictureResponse= pictureResponse;
}
}
.zipWith操作符。uploadMovies()和uploadPictures()返回Observable，uploadMovies()
.zipWith(uploadPictures()) { m, p ->
"$m with $p were uploaded"
}
.subscribe { print(it) }