在flatmap操作后访问map操作内的单对象

Question

exchange.getAttributes().put(GATEWAY_REQUEST_URL_ATTR, URI.create("https://" + newTargetURLHost + )

newTarget = serviceReturnsMono.getServerMapping(id).block().getHost();

Mono.just(serviceReturnsMono.getServerMapping(id))
                    .flatMap(flat -> flat)
                    .subscribeOn(Schedulers.immediate())
                    .map(server -> exchange.getAttributes().put(GATEWAY_REQUEST_URL_ATTR, URI.create("https://" + server.getHost() ))
                    .subscribe();

执行上述代码时，exchange属性不会发生变异。

我也尝试过以下方法，但没有成功：

            serverMappingMono
                    .map(serverMapping -> exchange.getAttributes().put(GATEWAY_REQUEST_URL_ATTR, URI.create("https://" + serverMapping.getHost() ))
                    .subscribe();

作为一个测试，当我按照下面的方式修改代码以排除故障时，下面的代码确实会发出一个硬编码字符串，并且exchange属性发生了变化。

                    .map(server -> exchange.getAttributes().put(GATEWAY_REQUEST_URL_ATTR, URI.create("https://" + "testHostName" ))
                    .subscribe();

    private ReturnsMonoServerMappingService returnsMonoServerMappingService;

    @Override
    public int getOrder() {
        return 10001;
    }

    @Override
    public Mono<Void> filter(ServerWebExchange exchange, GatewayFilterChain chain) {

        final String id = exchange.getRequest().getHeaders().getFirst("reference");

        return chain.filter(exchange).then(Mono.fromRunnable(() -> {

            Mono<ServerMapping> serverMapping = returnsMonoServerMappingService.getServerMapping(id);
            serverMapping
                    .map(server -> exchange.getAttributes().put(GATEWAY_REQUEST_URL_ATTR, URI.create("https://" + server.getHost()  )))
                    .subscribe();

        }));
    }
}

@Override
public Mono<Void> filter(ServerWebExchange exchange, GatewayFilterChain chain) {

    final String id = exchange.getRequest()
                                 .getHeaders()
                                 .getFirst("reference");

    return returnsMonoServerMappingService.getServerMapping(id)
                     .doOnSuccess(serverMapping -> {                
                         exchange.getAttributes()
                             .put(GATEWAY_REQUEST_URL_ATTR, URI.create("https://" + server.getHost()
        }).then(chain.filter(exchange));
}

Answer 1

我在我的手机上写这个，所以不能测试它，我从记忆中写，但它应该是这样的东西，我想。或者至少你明白了要旨。

我们首先提取ID。然后我们查找servermapping，如果进行顺利，我们将其作为一个属性，然后继续筛选器链。

您几乎不应该在应用程序中订阅，调用客户端通常是订阅方。并且永远不要在反应性应用程序中阻塞。

@Override
public Mono<Void> filter(ServerWebExchange exchange, GatewayFilterChain chain) {

    final String id = exchange.getRequest()
                                 .getHeaders()
                                 .getFirst("reference");
    
    return returnsMonoServerMappingService.getServerMapping(id)
                     .doOnSuccess(serverMapping -> {                
                         exchange.getAttributes()
                             .put(GATEWAY_REQUEST_URL_ATTR, URI.create("https://" + server.getHost()
        }).then(chain.filter(exchange));
}