37

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

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();
}

上面的代码适用于小有效载荷，但不适用于大有效载荷，我认为这是因为我只读取单个通量值next我不确定如何组合和读取所有dataBuffer.

我是反应堆的新手，所以我不知道很多关于通量/单声道的技巧。

4

6 回答 6

18

这实际上并不像其他答案所暗示的那么复杂。

正如@jin-kwon 所建议的那样，流式传输数据而不将其全部缓冲在内存中的唯一方法是使用管道。但是，它可以通过使用 Spring 的BodyExtractors和DataBufferUtils实用程序类非常简单地完成。

例子：

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为失败状态代码抛出异常，您可以使用跳过block()调用和中间ClientResponse变量

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

反而。

于 2019-10-07T14:50:46.003 回答
10

Bk Santiago 的回答稍作修改后使用了 ofreduce()而不是collect(). 非常相似，但不需要额外的类：

爪哇：

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 */ }

这种方法比使用的好处collect()是你不需要有不同的类来收集东西。

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

于 2018-08-14T15:22:20.043 回答
7

这是其他答案的另一个变体。而且它仍然对内存不友好。

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();
}
于 2019-10-13T09:42:14.620 回答
4

我能够通过使用Flux#collect和SequenceInputStream

@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;
  }
}
于 2017-09-28T06:33:05.857 回答
3

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，它违背了使用非阻塞 HTTP 客户端的目的，更不用说聚合整个响应了。有关Java NIO 与 IO 的比较，请参阅此内容。

于 2020-07-27T06:24:42.520 回答
1

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);
                     }
                 });
}
于 2019-05-12T05:36:56.837 回答