如何在SpringWebFlux中获取当前控制器方法的URL?

你可以使用 ServerHttpRequest 作为方法参数获取 uri :

@RestController
public class GreetingController {
  @GetMapping("/greeting")
  public Mono<Greeting> getGreeting(ServerHttpRequest serverHttpRequest) {
    return Mono.just(new Greeting("greeting", serverHttpRequest.getURI().toString()));
  }
}

更可取地 linkToCurrentResource 应该理解 X-Forwarded-??? 如果在负载平衡器后面运行,则返回头。

然后你可以暴露 ForwardedHeaderTransformer @Bean .

从其文档中:

从“”中提取值 Forwarded “和” X-Forwarded-* “要覆盖请求的邮件头 URI (即 HttpRequest.getURI() )所以它反映了客户机发起的协议和地址。

@Configuration
open class MvcConfig {
    @Bean
    open fun forwardedHeaderTransformer() = ForwardedHeaderTransformer()
}

以下是一些测试:

@ExtendWith(SpringExtension::class)
@SpringBootTest(webEnvironment = SpringBootTest.WebEnvironment.DEFINED_PORT,
        properties = ["server.port=4333"])
class GreetingController2Test {
    @Autowired
    private lateinit var restTemplate: TestRestTemplate

    @Test
    fun `should return uri`() {
        val responseEntity = restTemplate.getForEntity("/greeting", Greeting::class.java)
        val greeting = responseEntity.body!!
        assertEquals("http://localhost:4333/greeting", greeting.uri)
    }

    @Test
    fun `should return uri composed from forwarded-??? headers`() {
        val headers = HttpHeaders()
        headers["X-Forwarded-Host"] = "external-uri.com"
        headers["X-Forwarded-Proto"] = "https"
        headers["X-Forwarded-Prefix"] = "/prefix"

        val httpEntity = HttpEntity(null, headers)
        val responseEntity = restTemplate.exchange("/greeting", HttpMethod.GET, httpEntity, Greeting::class.java)
        val greeting = responseEntity.body!!
        assertEquals("https://external-uri.com/prefix/greeting", greeting.uri)
    }
}

以及 Greeting.kt :

data class Greeting(
        val g: String? = null,
        val uri: String? = null
)