类型与akka.http.scaladsl.server.Route不匹配 [英] type mismatch with akka.http.scaladsl.server.Route

问题描述

我已经使用akka http创建了一个http服务器，如下所示：

I've created a http server with akka http as follows:

import akka.actor.typed.{ActorRef, ActorSystem}
import akka.http.scaladsl.Http
import akka.http.scaladsl.server.Route
import com.sweetsoft.LoggerActor.Log
import akka.actor.typed.scaladsl.adapter._
import akka.http.scaladsl.Http.ServerBinding
import akka.http.scaladsl.model._
import com.sweetsoft._
import akka.http.scaladsl.server.Directives._
import akka.stream.typed.scaladsl.ActorMaterializer

import scala.concurrent.Future

object ProducerActor {

  private val route: Option[ActorRef[ProducerMessage]] => Option[ActorRef[Log]] => Route
  = store => logger =>
    path("producer") {
      post {
        entity(as[ProducerMessage]) { msg =>
          complete(HttpEntity(ContentTypes.`text/html(UTF-8)`, "<h1>Say hello to akka-http</h1>"))
        }
      }
    }

  def create[A](store: Option[ActorRef[ProducerMessage]], logger: Option[ActorRef[Log]])
               (implicit system: ActorSystem[A])
  : Future[ServerBinding] = {

    implicit val materializer = ActorMaterializer()

    //Please log
    Http()(system.toUntyped).bindAndHandle(route(store)(logger), getServerIp, getServerPort)
  }

}

编译器抱怨：

[error] /home/developer/scala/plugger/src/main/scala/com/sweetsoft/producer/ProducerActor.scala:35:56: type mismatch;
[error]  found   : akka.http.scaladsl.server.Route
[error]     (which expands to)  akka.http.scaladsl.server.RequestContext => scala.concurrent.Future[akka.http.scaladsl.server.RouteResult]
[error]  required: akka.stream.scaladsl.Flow[akka.http.scaladsl.model.HttpRequest,akka.http.scaladsl.model.HttpResponse,Any]
[error]     Http()(system.toUntyped).bindAndHandle(route(store)(logger), getServerIp, getServerPort)

我是否忘记导入任何库？

Do I forget to import any libraries?

推荐答案

从< a href = https://doc.akka.io/docs/akka-http/10.1.9/routing-dsl/routes.html rel = nofollow noreferrer>文档：

使用 Route.handlerFlow 或 Route.asyncHandler Route 可以提升为处理程序 Flow 或异步处理程序函数，以与 bindAndHandleXXX 。

Using Route.handlerFlow or Route.asyncHandler a Route can be lifted into a handler Flow or async handler function to be used with a bindAndHandleXXX call from the Core Server API.

注意： Route 到 RouteResu中定义的Flow [HttpRequest，HttpResponse，单位] lt 随播广告，它依赖于 Route.handlerFlow 。

因此，您至少有三个选择：

Therefore, you have at least three options:

1. 调用 Route.handlerFlow ：

1. Call Route.handlerFlow:

...bindAndHandle(Route.handlerFlow(route(store)(logger)), ...)

1. 导入方法在 Route 随播对象中，并执行与上述相同的操作，只不过现在您可以删除对 Route 对象的显式引用：

1. Import the methods in the Route companion object and do the same as above, except now you can drop the explicit reference to the Route object:

import akka.http.scaladsl.server.Route
import akka.http.scaladsl.server.Route._

...bindAndHandle(handlerFlow(route(store)(logger)), ...)

1. 导入 akka.http.scaladsl.server.RouteResult ._ ：

1. Import akka.http.scaladsl.server.RouteResult._:

import akka.http.scaladsl.server.RouteResult._

...bindAndHandle(route(store)(logger), ...)

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