未定义Scala案例类的备用构造函数:没有足够的方法参数

Question

我无法弄清楚为什么这不起作用...在编译期间我收到以下错误:

[error] /Users/zbeckman/Projects/Glimpulse/Server-2/project/glimpulse-server/app/service/GPGlimpleService.scala:17: not enough arguments for method apply: (id: Long, glimpleId: Long, layerOrder: Int, created: Long, attachments: List[models.GPAttachment])models.GPLayer in object GPLayer.
[error] Unspecified value parameter attachments.
[error]     private val layer1: List[GPLayer] = List(GPLayer(1, 42, 1, 9), GPLayer(2, 42, 2, 9))

对于这个案例类...请注意备用构造函数的定义:

case class GPLayer(id: Long, glimpleId: Long, layerOrder: Int, created: Long, attachments: List[GPAttachment]) {
    def this(id: Long, glimpleId: Long, layerOrder: Int, created: Long) = this(id, glimpleId, layerOrder, created, List[GPAttachment]())
}

Answer 1

GPLayer(1, 42, 1, 9)

和写作一样

GPLayer.apply(1, 42, 1, 9)

因此,您应该apply在随播对象中定义替代方法,而不是定义替代构造函数GPLayer.

case class GPLayer(id: Long, glimpleId: Long, layerOrder: Int, created: Long, attachments: List[GPAttachment]) 

object GPLayer {
  def apply(id: Long, glimpleId: Long, layerOrder: Int, created: Long) = GPLayer(id, glimpleId, layerOrder, created, List[GPAttachment]())
}

如果要改为调用altnernative构造函数,则必须添加new-keyword:

new GPLayer(1, 42, 1, 9)

编辑:正如Nicolas Cailloux所提到的,您的替代构造函数实际上只是为成员提供默认值attachments,因此最佳解决方案实际上是不引入新方法,而是指定此默认值,如下所示:

case class GPLayer(id: Long, glimpleId: Long, layerOrder: Int, created: Long, attachments: List[GPAttachment] = Nil)