噴霧：在性狀重寫功能

Question

我使用的噴霧，有一個基本路線特點，它定義了一些應該實現的功能...

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

    lazy val serviceBaseRoute = ... 

然後我MIXIN這種特質對其他一些性狀...

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 

最後我作爲構建遵循的路線的頂層小號...

trait V1Routes extends HttpService 
    with MyRoute1 
    with MyRoute2 { 

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

編譯罰款等......然而，當我運行這個MyRoute2的功能覆蓋覆蓋在MyRoute1定義。我在想這是因爲我沒有MyRoute1和MyRoute2的實際實例，因爲它們被簡單地混合在一起，並且自MyRoute2之後的MyRoute2的值覆蓋了MyRoute1的值之後呢？ 那麼實現這種保存每個特徵中定義的覆蓋的最佳方式是什麼？ 我是否需要將MyRoute1和MyRoute2定義爲對象？

感謝

Answer 1

根據您的意見，其目的是在MyRoute1和MyRoute2，其中每個被定製serviceBaseRoute創建不同的路線。由於ServiceBaseRoute由兩個功能定製，你可以做serviceBaseRoute高階函數，並通過這兩個函數的實現它的子性狀：

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 
     } 
    } 
}