我需要一些帮助来实现RxJava中的并行异步调用。我已经选择了一个简单的用例,其中FIRST调用获取(而不是搜索)要显示的产品列表(Tile)。随后的电话会出来并获取(A)评论和(B)产品图片
经过几次尝试,我到了这个地方。
1 Observable<Tile> searchTile = searchServiceClient.getSearchResults(searchTerm);
2 List<Tile> allTiles = new ArrayList<Tile>();
3 ClientResponse response = new ClientResponse();
4 searchTile.parallel(oTile -> {
5 return oTile.flatMap(t -> {
6 Observable<Reviews> reviews = reviewsServiceClient.getSellerReviews(t.getSellerId());
7 Observable<String> imageUrl = reviewsServiceClient.getProductImage(t.getProductId());
8 return Observable.zip(reviews, imageUrl, (r, u) -> {
9 t.setReviews(r);
10 t.setImageUrl(u);
11 return t;
12 });
13 });
14 }).subscribe(e -> {
15 allTiles.add((Tile) e);
16 });
第1行:熄灭并获取要显示的产品(平铺)
第4行:我们获取Observable和SHARD的列表以获取评论和imageUrls
谎言6,7:获取Observable评论和Observable url
第8行:最后压缩2个可观察量以返回更新的Observable
第15行:最后第15行整理所有要在集合中显示的单个产品,这些产品可以返回到调用层
虽然Observable已被分片,但在我们的测试中运行了4个不同的线程;获取评论和图像似乎是一个接一个。我怀疑第8行的zip步骤基本上导致了2个observables(review和url)的顺序调用。
这个小组是否有任何关于平行获取reiews和图片网址的建议。实质上,上面附带的瀑布图应该看起来更垂直堆叠。对评论和图片的调用应该是并行的
感谢 阿南德拉曼
答案 0 :(得分:87)
对于几乎所有用例,并行运算符都被证明是一个问题,并且没有达到预期的效果,因此在1.0.0.rc.4版本中将其删除:https://github.com/ReactiveX/RxJava/pull/1716
可以看到如何执行此类行为并获得并行执行的一个很好的示例{。{3}}。
在您的示例代码中,不清楚searchServiceClient
是同步还是异步。它会影响如何稍微解决问题,就像它已经是异步一样,不需要额外的调度。如果需要同步额外调度。
首先是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 {
}
}
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
我已经模拟了每个IO调用以获取1000毫秒,因此很明显延迟在哪里并且它是并行发生的。它以经过的毫秒打印出进度。
这里的技巧是flatMap合并异步调用,因此只要合并的Observable是异步的,它们就会同时执行。
如果getProductImage(t.getProductId())
这样的调用是同步的,可以像这样调用异步:getProductImage(t.getProductId())。subscribeOn(Schedulers.io)。
以上是没有所有日志记录和样板文件类型的上述示例的重要部分:
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();
我希望这会有所帮助。
答案 1 :(得分:4)
仍然是@ JDK 7的人,他们的IDE还没有自动检测JDK 8源代码,以及如何通过@benjchristensen尝试上述出色的响应(和解释),可以使用这种无耻折射的JDK 7,码。感谢@benjchristensen提供了一个惊人的解释和示例!
import java.util.List;
import rx.Observable;
import rx.Subscriber;
import rx.functions.Action0;
import rx.functions.Func1;
import rx.functions.Func2;
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(new Action0()
{
@Override
public void call()
{
logTime("Search started ", startTime);
}
})
.doOnCompleted(new Action0()
{
@Override
public void call()
{
logTime("Search completed ", startTime);
}
});
Observable<TileResponse> populatedTiles = searchTile.flatMap(new Func1<Tile, Observable<TileResponse>>()
{
@Override
public Observable<TileResponse> call(final Tile t)
{
Observable<Reviews> reviews = getSellerReviews(t.getSellerId())
.doOnCompleted(new Action0()
{
@Override
public void call()
{
logTime("getSellerReviews[" + t.id + "] completed ", startTime);
}
});
Observable<String> imageUrl = getProductImage(t.getProductId())
.doOnCompleted(new Action0()
{
@Override
public void call()
{
logTime("getProductImage[" + t.id + "] completed ", startTime);
}
});
return Observable.zip(reviews, imageUrl, new Func2<Reviews, String, TileResponse>()
{
@Override
public TileResponse call(Reviews r, String u)
{
return new TileResponse(t, r, u);
}
})
.doOnCompleted(new Action0()
{
@Override
public void call()
{
logTime("zip[" + t.id + "] completed ", startTime);
}
});
}
});
List<TileResponse> allTiles = populatedTiles
.toList()
.doOnCompleted(new Action0()
{
@Override
public void call()
{
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(final T... ts)
{
return Observable.create(new Observable.OnSubscribe<T>()
{
@Override
public void call(Subscriber<? super T> s)
{
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
{
}
}