并行运算符被证明是几乎所有用例的问题，并且没有达到大多数人期望的效果，因此它在 1.0.0.rc.4 版本中被删除：https://github.com/ReactiveX/RxJava/pull/1716
here 有一个很好的例子来说明如何执行这种类型的行为并获得并行执行。
在您的示例代码中，不清楚searchServiceClient 是同步的还是异步的。它会稍微影响如何解决问题，就好像它已经是异步的，不需要额外的调度。如果需要同步额外调度。
首先这里有一些显示同步和异步行为的简单示例：
import rx.Observable;
import rx.Subscriber;
import rx.schedulers.Schedulers;
public class ParallelExecution {
public static void main(String[] args) {
System.out.println("------------ mergingAsync");
mergingAsync();
System.out.println("------------ mergingSync");
mergingSync();
System.out.println("------------ mergingSyncMadeAsync");
mergingSyncMadeAsync();
System.out.println("------------ flatMapExampleSync");
flatMapExampleSync();
System.out.println("------------ flatMapExampleAsync");
flatMapExampleAsync();
System.out.println("------------");
}
private static void mergingAsync() {
Observable.merge(getDataAsync(1), getDataAsync(2)).toBlocking().forEach(System.out::println);
}
private static void mergingSync() {
// here you'll see the delay as each is executed synchronously
Observable.merge(getDataSync(1), getDataSync(2)).toBlocking().forEach(System.out::println);
}
private static void mergingSyncMadeAsync() {
// if you have something synchronous and want to make it async, you can schedule it like this
// so here we see both executed concurrently
Observable.merge(getDataSync(1).subscribeOn(Schedulers.io()), getDataSync(2).subscribeOn(Schedulers.io())).toBlocking().forEach(System.out::println);
}
private static void flatMapExampleAsync() {
Observable.range(0, 5).flatMap(i -> {
return getDataAsync(i);
}).toBlocking().forEach(System.out::println);
}
private static void flatMapExampleSync() {
Observable.range(0, 5).flatMap(i -> {
return getDataSync(i);
}).toBlocking().forEach(System.out::println);
}
// artificial representations of IO work
static Observable<Integer> getDataAsync(int i) {
return getDataSync(i).subscribeOn(Schedulers.io());
}
static Observable<Integer> getDataSync(int i) {
return Observable.create((Subscriber<? super Integer> s) -> {
// simulate latency
try {
Thread.sleep(1000);
} catch (Exception e) {
e.printStackTrace();
}
s.onNext(i);
s.onCompleted();
});
}
}
以下是尝试提供与您的代码更匹配的示例：
import java.util.List;
import rx.Observable;
import rx.Subscriber;
import rx.schedulers.Schedulers;
public class ParallelExecutionExample {
public static void main(String[] args) {
final long startTime = System.currentTimeMillis();
Observable<Tile> searchTile = getSearchResults("search term")
.doOnSubscribe(() -> logTime("Search started ", startTime))
.doOnCompleted(() -> logTime("Search completed ", startTime));
Observable<TileResponse> populatedTiles = searchTile.flatMap(t -> {
Observable<Reviews> reviews = getSellerReviews(t.getSellerId())
.doOnCompleted(() -> logTime("getSellerReviews[" + t.id + "] completed ", startTime));
Observable<String> imageUrl = getProductImage(t.getProductId())
.doOnCompleted(() -> logTime("getProductImage[" + t.id + "] completed ", startTime));
return Observable.zip(reviews, imageUrl, (r, u) -> {
return new TileResponse(t, r, u);
}).doOnCompleted(() -> logTime("zip[" + t.id + "] completed ", startTime));
});
List<TileResponse> allTiles = populatedTiles.toList()
.doOnCompleted(() -> logTime("All Tiles Completed ", startTime))
.toBlocking().single();
}
private static Observable<Tile> getSearchResults(String string) {
return mockClient(new Tile(1), new Tile(2), new Tile(3));
}
private static Observable<Reviews> getSellerReviews(int id) {
return mockClient(new Reviews());
}
private static Observable<String> getProductImage(int id) {
return mockClient("image_" + id);
}
private static void logTime(String message, long startTime) {
System.out.println(message + " => " + (System.currentTimeMillis() - startTime) + "ms");
}
private static <T> Observable<T> mockClient(T... ts) {
return Observable.create((Subscriber<? super T> s) -> {
// simulate latency
try {
Thread.sleep(1000);
} catch (Exception e) {
}
for (T t : ts) {
s.onNext(t);
}
s.onCompleted();
}).subscribeOn(Schedulers.io());
// note the use of subscribeOn to make an otherwise synchronous Observable async
}
public static class TileResponse {
public TileResponse(Tile t, Reviews r, String u) {
// store the values
}
}
public static class Tile {
private final int id;
public Tile(int i) {
this.id = i;
}
public int getSellerId() {
return id;
}
public int getProductId() {
return id;
}
}
public static class Reviews {
}
}
这个输出：
Search started => 65ms
Search completed => 1094ms
getProductImage[1] completed => 2095ms
getSellerReviews[2] completed => 2095ms
getProductImage[3] completed => 2095ms
zip[1] completed => 2096ms
zip[2] completed => 2096ms
getProductImage[2] completed => 2096ms
getSellerReviews[1] completed => 2096ms
zip[3] completed => 2096ms
All Tiles Completed => 2097ms
getSellerReviews[3] completed => 2097ms
我已将每个 IO 调用模拟为花费 1000 毫秒，因此很明显延迟在哪里并且它是并行发生的。它以经过的毫秒数打印出进度。
这里的技巧是 flatMap 会合并异步调用，所以只要被合并的 Observable 是异步的，它们都会被并发执行。
如果像getProductImage(t.getProductId()) 这样的调用是同步的，则可以像这样使其异步：getProductImage(t.getProductId()).subscribeOn(Schedulers.io)。
这是上面示例的重要部分，没有所有的日志记录和样板类型：
Observable<Tile> searchTile = getSearchResults("search term");;
Observable<TileResponse> populatedTiles = searchTile.flatMap(t -> {
Observable<Reviews> reviews = getSellerReviews(t.getSellerId());
Observable<String> imageUrl = getProductImage(t.getProductId());
return Observable.zip(reviews, imageUrl, (r, u) -> {
return new TileResponse(t, r, u);
});
});
List<TileResponse> allTiles = populatedTiles.toList()
.toBlocking().single();
我希望这会有所帮助。