使用示例ReactorNettyWebSocketClient

Ale*_*zel 4 spring-boot spring-websocket spring-webflux

New Spring在Spring文档中有一些WebSocketClient示例。

WebSocketClient client = new ReactorNettyWebSocketClient();
client.execute("ws://localhost:8080/echo"), session -> {... }).blockMillis(5000);
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但这很短,也不清楚:

  1. 如何向服务器(订阅频道)发送消息?
  2. 然后处理传入的流并发出Flux消息?
  3. 连接中断时重新连接到服务器?

有人可以提供更复杂的例子吗?

UPD。我试图做类似的事情:

public Flux<String> getStreaming() {

    WebSocketClient client = new ReactorNettyWebSocketClient();
    EmitterProcessor<String> output = EmitterProcessor.create();
    Flux<String> input = Flux.just("{ event: 'subscribe', channel: 'examplpe' }");

    Mono<Void> sessionMono = client.execute(URI.create("ws://api.example.com/"),
            session -> session
                    .send(input.map(session::textMessage))
                    .thenMany(session.receive().map(WebSocketMessage::getPayloadAsText).subscribeWith(output).then())
                    .then());

    return output.doOnSubscribe(s -> sessionMono.subscribe());
}
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但这仅返回一条消息。就像我没有订阅。

cri*_*ixx 5

我假设您正在使用“回声”服务。为了从服务中获取一些消息,您必须将它们推入websocket,服务将“回显”它们给您。

在示例代码中,您仅向websocket写入了一个元素。一旦您将更多消息推入套接字,您将获得更多回报。

我修改了代码以连接到ws://echo.websocket.org本地服务。浏览到/stream每一秒时,都会显示一条新消息。

@GetMapping(path = "/stream", produces = MediaType.TEXT_EVENT_STREAM_VALUE)
public Flux<String> getStreaming() throws URISyntaxException {

    Flux<String> input = Flux.<String>generate(sink -> sink.next(String.format("{ message: 'got message', date: '%s' }", new Date())))
        .delayElements(Duration.ofSeconds(1));

    WebSocketClient client = new ReactorNettyWebSocketClient();
    EmitterProcessor<String> output = EmitterProcessor.create();

    Mono<Void> sessionMono = client.execute(URI.create("ws://echo.websocket.org"), session -> session.send(input.map(session::textMessage))
        .thenMany(session.receive().map(WebSocketMessage::getPayloadAsText).subscribeWith(output).then()).then());

    return output.doOnSubscribe(s -> sessionMono.subscribe());
}
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希望这可以帮助...