如何在Spring Webflux中获取当前控制器方法的URL？

Question

这个问题类似于 What's the best way to get the current URL in Spring MVC? but for Webflux (it might also be related to 但这个问题更具体所以我希望有一个解决方案）。

我想做的是在 Webflux 控制器中获取当前 URL。例如：

@GetMapping("/greeting")
public Mono<Greeting> greeting() {
   String linkToCurrentResource = .. // How do I get this?
   Greeting greeting = new Greeting("hello",  linkToCurrentResource);
   return Mono.just(greeting);
}

最好linkToCurrentResource 应该理解X-Forwarded-??? headers 如果运行在负载均衡器后面。所以在这种情况下，我希望 linkToCurrentResource 类似于 https://external-uri.com/api/greeting。如果重要的话，我正在使用 Netty 服务器实现。

Answer 1

您可以使用 ServerHttpRequest 作为方法参数来获取 uri:

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

Preferably linkToCurrentResource should understand X-Forwarded-??? headers if running behind a load balancer.

那你可以暴露一个ForwardedHeaderTransformer @Bean.

来自其文档：

Extract values from "Forwarded" and "X-Forwarded-*" headers to override the request URI (i.e. HttpRequest.getURI()) so it reflects the client-originated protocol and address.

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

这里有一些测试：

@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)
    }
}

和 Greeting.kt:

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

如何在Spring Webflux中获取当前控制器方法的URL？

How to get URL of the current controller method in Spring Webflux?

java

spring-boot

spring-webflux