如何正确读取 Flux<DataBuffer> 并将其转换为单个 inputStream

Question

我正在为我的 spring-boot 应用程序使用 WebClient 和自定义 BodyExtractorclass

WebClient webLCient = WebClient.create();
webClient.get()
   .uri(url, params)
   .accept(MediaType.APPLICATION.XML)
   .exchange()
   .flatMap(response -> {
     return response.body(new BodyExtractor());
   })

BodyExtractor.java

@Override
public Mono<T> extract(ClientHttpResponse response, BodyExtractor.Context context) {
  Flux<DataBuffer> body = response.getBody();
  body.map(dataBuffer -> {
    try {
      JaxBContext jc = JaxBContext.newInstance(SomeClass.class);
      Unmarshaller unmarshaller = jc.createUnmarshaller();

      return (T) unmarshaller.unmarshal(dataBuffer.asInputStream())
    } catch(Exception e){
       return null;
    }
  }).next();
}

以上代码适用于小负载但不适用于大负载，我认为这是因为我只使用 next 读取单个通量值并且我不确定如何组合和读取所有 dataBuffer.

我是 reactor 的新手，所以我不知道 flux/mono 的很多技巧。

Answer 1

我能够通过使用 Flux#collect 和 SequenceInputStream

使其工作

@Override
public Mono<T> extract(ClientHttpResponse response, BodyExtractor.Context context) {
  Flux<DataBuffer> body = response.getBody();
  return body.collect(InputStreamCollector::new, (t, dataBuffer)-> t.collectInputStream(dataBuffer.asInputStream))
    .map(inputStream -> {
      try {
        JaxBContext jc = JaxBContext.newInstance(SomeClass.class);
        Unmarshaller unmarshaller = jc.createUnmarshaller();

        return (T) unmarshaller.unmarshal(inputStream);
      } catch(Exception e){
        return null;
      }
  }).next();
}

InputStreamCollector.java

public class InputStreamCollector {
  private InputStream is;

  public void collectInputStream(InputStream is) {
    if (this.is == null) this.is = is;
    this.is = new SequenceInputStream(this.is, is);
  }

  public InputStream getInputStream() {
    return this.is;
  }
}

Answer 2

Bk Santiago 的答案稍作修改后使用 reduce() 而不是 collect()。非常相似，但不需要额外的 class:

Java:

body.reduce(new InputStream() {
    public int read() { return -1; }
  }, (s: InputStream, d: DataBuffer) -> new SequenceInputStream(s, d.asInputStream())
).flatMap(inputStream -> /* do something with single InputStream */

或科特林：

body.reduce(object : InputStream() {
  override fun read() = -1
}) { s: InputStream, d -> SequenceInputStream(s, d.asInputStream()) }
  .flatMap { inputStream -> /* do something with single InputStream */ }

与使用 collect() 相比，这种方法的好处是您不需要使用不同的 class 来收集东西。

我创建了一个新的空 InputStream()，但如果该语法令人困惑，您也可以将其替换为 ByteArrayInputStream("".toByteArray())，而不是创建一个空 ByteArrayInputStream 作为您的初始值。

Answer 3

您可以使用管道。

static <R> Mono<R> pipeAndApply(
        final Publisher<DataBuffer> source, final Executor executor,
        final Function<? super ReadableByteChannel, ? extends R> function) {
    return using(Pipe::open,
                 p -> {
                     executor.execute(() -> write(source, p.sink())
                             .doFinally(s -> {
                                 try {
                                     p.sink().close();
                                 } catch (final IOException ioe) {
                                     log.error("failed to close pipe.sink", ioe);
                                     throw new RuntimeException(ioe);
                                 }
                             })
                             .subscribe(releaseConsumer()));
                     return just(function.apply(p.source()));
                 },
                 p -> {
                     try {
                         p.source().close();
                     } catch (final IOException ioe) {
                         log.error("failed to close pipe.source", ioe);
                         throw new RuntimeException(ioe);
                     }
                 });
}

或使用CompletableFuture、

static <R> Mono<R> pipeAndApply(
        final Publisher<DataBuffer> source,
        final Function<? super ReadableByteChannel, ? extends R> function) {
    return using(Pipe::open,
                 p -> fromFuture(supplyAsync(() -> function.apply(p.source())))
                         .doFirst(() -> write(source, p.sink())
                                 .doFinally(s -> {
                                     try {
                                         p.sink().close();
                                     } catch (final IOException ioe) {
                                         log.error("failed to close pipe.sink", ioe);
                                         throw new RuntimeException(ioe);
                                     }
                                 })
                                 .subscribe(releaseConsumer())),
                 p -> {
                     try {
                         p.source().close();
                     } catch (final IOException ioe) {
                         log.error("failed to close pipe.source", ioe);
                         throw new RuntimeException(ioe);
                     }
                 });
}

Answer 4

这真的没有其他答案暗示的那么复杂。

正如@jin-kwon 所建议的那样，在不将数据全部缓冲在内存中的情况下流式传输数据的唯一方法是使用管道。但是，可以通过使用 Spring 的 BodyExtractors and DataBufferUtils 实用程序类.

非常简单地完成

示例：

private InputStream readAsInputStream(String url) throws IOException {
    PipedOutputStream osPipe = new PipedOutputStream();
    PipedInputStream isPipe = new PipedInputStream(osPipe);

    ClientResponse response = webClient.get().uri(url)
        .accept(MediaType.APPLICATION.XML)
        .exchange()
        .block();
    final int statusCode = response.rawStatusCode();
    // check HTTP status code, can throw exception if needed
    // ....

    Flux<DataBuffer> body = response.body(BodyExtractors.toDataBuffers())
        .doOnError(t -> {
            log.error("Error reading body.", t);
            // close pipe to force InputStream to error,
            // otherwise the returned InputStream will hang forever if an error occurs
            try(isPipe) {
              //no-op
            } catch (IOException ioe) {
                log.error("Error closing streams", ioe);
            }
        })
        .doFinally(s -> {
            try(osPipe) {
              //no-op
            } catch (IOException ioe) {
                log.error("Error closing streams", ioe);
            }
        });

    DataBufferUtils.write(body, osPipe)
        .subscribe(DataBufferUtils.releaseConsumer());

    return isPipe;
}

如果您不关心检查响应代码或抛出失败状态代码的异常，您可以使用 [=15= 跳过 block() 调用和中间 ClientResponse 变量]

flatMap(r -> r.body(BodyExtractors.toDataBuffers()))

相反。

Answer 5

这是其他答案的另一种变体。而且它仍然不利于记忆。

static Mono<InputStream> asStream(WebClient.ResponseSpec response) {
    return response.bodyToFlux(DataBuffer.class)
        .map(b -> b.asInputStream(true))
        .reduce(SequenceInputStream::new);
}

static void doSome(WebClient.ResponseSpec response) {
    asStream(response)
        .doOnNext(stream -> {
            // do some with stream
            // close the stream!!!
        })
        .block();
}

Answer 6

有一种更简洁的方法可以直接使用底层的 reactor-netty HttpClient，而不是使用 WebClient。组合层次结构是这样的：

WebClient -uses-> HttpClient -uses-> TcpClient

显示代码比解释更容易：

HttpClient.create()
    .get()
    .responseContent() // ByteBufFlux
    .aggregate() // ByteBufMono
    .asInputStream() // Mono<InputStream>
    .block() // We got an InputStream, yay!

但是，正如我已经指出的那样，使用 InputStream 是一种阻塞操作，这违背了使用非阻塞 HTTP 客户端的目的，更不用说聚合整个响应了。有关 Java NIO 与 IO 比较，请参阅 this。

如何正确读取 Flux<DataBuffer> 并将其转换为单个 inputStream

How to correctly read Flux<DataBuffer> and convert it to a single inputStream

java

spring

project-reactor

reactive-streams

spring-webflux