Question

我使用Spray并拥有一个Base Route特征，它定义了许多应该实现的功能......

trait ServiceBaseRoute extends HttpService {
  def function1():Type
  def function2():Type

  lazy val serviceBaseRoute = ...

然后我将这个特性混合到许多其他特征......

trait MyRoute1 extends HttpService
  with ServiceBaseRoute {
  override def function1():Type = {...}
  override def function2():Type = {...}

  val myRoute1 = serviceBaseRoute

和...

trait MyRoute2 extends HttpService
  with ServiceBaseRoute {
  override def function1():Type = {...}
  override def function2():Type = {...}

  val myRoute2 = serviceBaseRoute

最后，我按照以下方式构建路线的顶层......

trait V1Routes extends HttpService
  with MyRoute1
  with MyRoute2 {

  val v1Routes = 
    pathPrefix("v1") {
      authenticate(...) {
        myRoute1 ~ myRoute2
      }
    }

编译精细等...但是当我运行它时，MyRoute2的函数覆盖会覆盖MyRoute1中定义的那些。我想这是因为我没有MyRoute1和MyRoute2的实际实例，因为它们只是混合进来，因为在MyRoute1之后添加了MyRoute2，它的值会覆盖MyRoute1的值？那么实现这个的最佳方法是保留每个特征中定义的覆盖？我是否需要将MyRoute1和MyRoute2定义为对象？

由于

Answer 1

根据您的评论，目标是在MyRoute1和MyRoute2中设置不同的路由，其中每个路由都是通过自定义serviceBaseRoute创建的。由于ServiceBaseRoute是由两个函数自定义的，因此您可以使serviceBaseRoute成为更高阶函数，并将这两个函数的实现传递给子特征：

trait ServiceBaseRoute extends HttpService {

  def serviceBaseRoute(f1: () => Type, f2: () => Type)= ???
}

trait MyRoute1 extends HttpService
with ServiceBaseRoute {

  def f1ImplA(): Type = ???

  def f2ImplA(): Type = ???

  val myRoute1 = serviceBaseRoute(f1ImplA, f2ImplA)
}

trait MyRoute2 extends HttpService
with ServiceBaseRoute {

  def f1ImplB(): Type = ???

  def f2ImplB(): Type = ???

  val myRoute2 = serviceBaseRoute(f1ImplB, f2ImplB)
}

trait V1Routes extends HttpService
with MyRoute1
with MyRoute2 {

  val v1Routes =
    pathPrefix("v1") {
      authenticate(...)
      {
        myRoute1 ~ myRoute2
      }
    }
}

喷涂：在特征中覆盖功能

1 个答案: