java - 如何将基本 URI 设置为 Spring WebClient

Question

我正在尝试创建一个 Spring WebClient 来调用 REST API。为此，我编写了以下代码。

@Bean
public WebClient defaultWebClient() {

    var tcpClient = TcpClient.create()
            .option(ChannelOption.CONNECT_TIMEOUT_MILLIS, 3_000)
            .doOnConnected(connection ->
                    connection.addHandlerLast(new ReadTimeoutHandler(3))
                            .addHandlerLast(new WriteTimeoutHandler(3)));

    return WebClient.builder()
            .baseUrl("https://test.com")
            .clientConnector(new ReactorClientHttpConnector(HttpClient.from(tcpClient)))
            .defaultHeader(HttpHeaders.ACCEPT, "application/vnd.api+json")
            .build();
}

然后是带有以下代码的实际 URI。

String user = webClient
                .get()
                .uri(URI.create("/api/v1/user/1"))
                .header("Authorization", TOKEN)
                .header(HttpHeaders.ACCEPT, "application/json")
                .retrieve()
                .bodyToMono(String.class)
                .block();

但是当上面的代码执行时，它会抛出以下异常。

{"timestamp":"1608209276546","level":"ERROR","thread":"http-nio-8080-exec-1","mdc":{"tenantId":null},"logger":"org.apache.catalina.core.ContainerBase.[Tomcat].[localhost].[/].[dispatcherServlet]","message":"Servlet.service() for servlet [dispatcherServlet] in context with path [] threw exception [Request processing failed; nested exception is org.springframework.web.reactive.function.client.WebClientRequestException: Connection refused: no further information: /127.0.0.1:12012; nested exception is io.netty.channel.AbstractChannel$AnnotatedConnectException: Connection refused: no further information: /127.0.0.1:12012] with root cause","context":"default","exception":"java.net.ConnectException: Connection refused: no further information\r\n\tat java.base/sun.nio.ch.SocketChannelImpl.checkConnect(Native Method)\r\n\tat

我不确定我需要在这里更改哪里，以便它开始指向基本 URI 而不是本地主机 (127.0.0.1:12012)。另一件事是，如果我在 API 调用中传递了完全限定名称 ( https://test.com/api/v1/user/1)，那么它就可以正常工作。

Answer 1

来自 javadoc，方法 uri(URI uri)

Specify the URI using an absolute, fully constructed URI

所以这个版本的方法不使用配置中的基本 url。

相反，您应该使用 UriBuilder 指定 URI，它被注入(inject)到方法 uri(Function<UriBuilder,URI> uriFunction) 中.

String user = webClient
            .get()
            .uri(uriBuilder -> uriBuilder.path("/api/v1/user/1").build())
            ...