的并行操作被证明是几乎所有的用例问题，没有做什么最期待它，所以它被移除1.0.0.rc.4版本：https://github.com/ReactiveX/RxJava/pull/1716
一个很好的例子是如何做到这种行为并获得并行执行，可以看到here。
在您的示例代码，目前还不清楚searchServiceClient是同步或异步的。它会影响如何解决问题，就好像它已经是异步的，不需要额外的调度。如果需要同步的额外调度。
首先这里是显示同步和异步行为的一些简单的例子：
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();
});
}
}
以下是在提供更符合您的代码示例尝试：
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 {
}
}
此输出：
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
我做了每个IO调用模拟采取1000毫秒，所以我t在延迟和并行发生时很明显。它打印出进度是在经过的毫秒中进行的。
这里的技巧是flatMap合并异步调用，所以只要合并的Observables是异步的，它们将全部同时执行。
如果像getProductImage(t.getProductId())这样的调用是同步的，则可以使其成为异步，如下所示：getProductImage（t.getProductId（））。subscribeOn（Schedulers.io）。
这里是上面的例子中没有所有的记录和样板类型的重要组成部分：
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();
我希望这有助于。
您是如何生成Transfer Timeline图表的？它看起来很酷很有用。想自己使用它。 – 2015-01-18 02:24:04
由于我的系统正在进行外部呼叫，我只是通过提琴手代理呼叫。提琴手可以选择生成网络时间表。你基本上看到了这个观点。在为代理请求设置提琴手之后;只需选择您感兴趣的会话，然后单击右侧窗格中的时间轴选项卡即可。谢谢anand – diduknow 2015-01-19 07:52:11