在Scala案例类上未定义的替代构造函数:方法不足

Question

我无法弄清楚为什么这不起作用...在编译期间，我会收到以下错误：

[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))

对于此情况类别...请注意替代构造函数的定义：

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)

与写作相同

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

因此，您应该在伴随对象GPLayer中定义替代的apply方法，而不是定义替代构造函数。

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]())
}

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

new GPLayer(1, 42, 1, 9)

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编辑：正如Nicolas Cailloux所述，您的替代构造函数实际上只是为成员attachments提供默认值，因此最好的解决方案实际上是不引入新方法，但要指定此方法默认值如下：

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

Answer 2

请注意，在您的情况下，您可以为最后一个参数提供默认值：

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