akka http使用Json支持和xmlsupport

Question

我们能同时使用spray-json支持https://doc.akka.io/docs/akka-http/current/common/json-support.html和XML支持https://doc.akka.io/docs/akka-http/current/common/xml-support.html吗

 private[rest] def route =
  (pathPrefix("employee") & get) {
    path(Segment) { id =>
      parameters('department ? "") { (flag) =>
        extractUri { uri =>
          complete {
            flag match {
              case "hr": =>  {

            HttpEntity(MediaTypes.`application/xml`.withCharset(HttpCharsets.`UTF-8`),"hr department")
            }
              case "tech" =>{
                HttpEntity(ContentType(MediaTypes.`application/json`), mapper.writeValueAsString("tech department"))

              }

            }
          }
        }
      }
    }
  }

我尝试了以下方法，通过使用JsonProtocols和ScalaXmlSupport，我得到了预期的ResponseMarshallable的编译错误，但找到了Department

case class department(name:String)
private[rest] def route =
  (pathPrefix("employee") & get) {
    path(Segment) { id =>
      parameters('department ? "") { (flag) =>
        extractUri { uri =>
          complete {
            flag match {
              case "hr": =>  {

            complete(department(name =flag))
            }
              case "tech" =>{
                complete(department(name =flag))

              }

            }
          }
        }
      }
    }
  }

Answer 1

我认为有几个问题你必须克服以达到你想要的：

>

您希望根据请求参数自定义响应类型。这意味着基于标准隐式的封送处理对您不起作用，您必须执行一些显式步骤

您希望将一些业务对象封送到XML字符串中。不幸的是，您引用的scalaxmlsupport不支持这一点，它只能将XML树封送到响应中。所以您需要一些可以执行XML序列化的库。一个选择是将jackson-dataformat-xml与jackson-module-scala一起使用。这还意味着您必须编写自己的自定义封送器。幸运的是没有那么难。

import akka.http.scaladsl.marshalling.{ToResponseMarshallable, Marshaller}

// json marshalling
import akka.http.scaladsl.marshallers.sprayjson.SprayJsonSupport
import spray.json._
import spray.json.DefaultJsonProtocol._
implicit val departmentFormat = DefaultJsonProtocol.jsonFormat1(department)
val departmentJsonMarshaller = SprayJsonSupport.sprayJsonMarshaller[department]

// xml marshalling, need to write a custom Marshaller
// there are several MediaTypes for XML such as `application/xml` and `text/xml`, you'll have to choose the one you need.
val departmentXmlMarshaller = Marshaller.StringMarshaller.wrap(MediaTypes.`application/xml`)((d: department) => {
  import com.fasterxml.jackson.dataformat.xml.XmlMapper
  import com.fasterxml.jackson.module.scala.DefaultScalaModule
  val mapper = new XmlMapper()
  mapper.registerModule(DefaultScalaModule)
  mapper.writeValueAsString(d)
})


private val route =
  (pathPrefix("employee") & get) {
    path(Segment) { id =>
      parameters('department ? "") { (flag) =>
        extractUri { uri => {
          flag match {
            case "hr" => {
              // explicitly use the XML marshaller 
              complete(ToResponseMarshallable(department(name = flag))(departmentXmlMarshaller))
            }
            case "tech" => {
              // explicitly use the JSON marshaller 
              complete(ToResponseMarshallable(department(name = flag))(departmentJsonMarshaller))
            }
          }
        }
        }
      }
    }
  }