Scala:如何为基本的父子关系建模 [英] scala: how to model a basic parent-child relation

问题描述

我有一个品牌类，其中有几种产品

I have a Brand class that has several products

在产品类别中，我想引用该品牌，例如:

And in the product class I want to have a reference to the brand, like this:

case class Brand(val name:String, val products: List[Product])

case class Product(val name: String, val brand: Brand)

我如何填写这些课程?

我的意思是，除非拥有品牌，否则我无法创建产品

I mean, I can't create a product unless I have a brand

除非拥有产品列表，否则我无法创建品牌(因为Brand.products是有效值)

And I can't create the brand unless I have a list of Products (because Brand.products is a val)

为这种关系建模的最佳方法是什么?

What would be the best way to model this kind of relation?

推荐答案

我会问您为什么要重复这些信息，方法是说列表和每种产品中哪些产品与哪个品牌相关.

I would question why you are repeating the information, by saying which products relate to which brand in both the List and in each Product.

仍然可以做到:

class Brand(val name: String, ps: => List[Product]) {
  lazy val products = ps
  override def toString = "Brand("+name+", "+products+")" 
}

class Product(val name: String, b: => Brand) { 
  lazy val brand = b
  override def toString = "Product("+name+", "+brand.name+")"
}

lazy val p1: Product = new Product("fish", birdseye)
lazy val p2: Product = new Product("peas", birdseye)
lazy val birdseye = new Brand("BirdsEye", List(p1, p2))

println(birdseye) 
  //Brand(BirdsEye, List(Product(fish, BirdsEye), Product(peas, BirdsEye)))

不幸的是，案例类似乎不允许使用按名称命名的参数.

By-name params don't seem to be allowed for case classes unfortunately.

另请参阅类似的问题:实例化不可变的配对对象

See also this similar question: Instantiating immutable paired objects

这篇关于Scala:如何为基本的父子关系建模的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！