问题
I have a Brand class that has several products
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)
How can I poulate these classes???
I mean, I can't create a product unless I have a brand
And I can't create the brand unless I have a list of Products (because Brand.products is a val)
What would be the best way to model this kind of relation?
回答1:
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.
Still, you can do it:
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)))
By-name params don't seem to be allowed for case classes unfortunately.
See also this similar question: Instantiating immutable paired objects
回答2:
Since your question is about model to this relationship, I will say why not just model them like what we do in database? Separate the entity and the relationship.
val productsOfBrand: Map[Brand, List[Product]] = {
// Initial your brand to products mapping here, using var
// or mutable map to construct the relation is fine, since
// it is limit to this scope, and transparent to the outside
// world
}
case class Brand(val name:String){
def products = productsOfBrand.get(this).getOrElse(Nil)
}
case class Product(val name: String, val brand: Brand) // If you really need that brand reference
来源:https://stackoverflow.com/questions/10262574/scala-how-to-model-a-basic-parent-child-relation