Scala:如何从某个集合创建XML节点

如果你有类似的东西:

val myStuff = Array(Person("joe",40), Person("mary", 35))

如何使用该数据创建XML值作为节点？我知道如何在XML表达式中使用{braces}来放置值,但这是一组值.我需要明确迭代还是有更好的东西？

val myXml = { /* what here?! */ }

结果值应该是这样的:

joe40
mary39

Aaron Maenpa.. 21

由于它是一种函数式编程语言,Array.map可能正是您所寻找的:

class Person(name : String, age : Int){
    def toXml() = { name }{ age }
}

object xml {
    val people = List(
        new Person("Alice", 16),
        new Person("Bob", 64)
    )

    val data = { people.map(p => p.toXml()) }

    def main(args : Array[String]){
        println(data)
    }
}

结果是:

Alice16Bob64

格式化结果(为了更好的阅读):

   
      Alice
      16
   
   
      Bob
      64
   

hishadow.. 5

为完整起见,您还可以使用for..yield(或函数调用):

import scala.xml

case class Person(val name: String, val age: Int) {
  def toXml(): xml.Elem =
    { name }{ age }
}

def peopleToXml(people: List[Person]): xml.Elem = {
  {
    for {person <- people if person.age > 39}
      yield person.toXml
  }
}

val data = List(Person("joe",40),Person("mary", 35))
println(peopleToXml(data))

(Woody Folsom指出的固定误差)

由于它是一种函数式编程语言,Array.map可能正是您所寻找的:

class Person(name : String, age : Int){
    def toXml() = { name }{ age }
}

object xml {
    val people = List(
        new Person("Alice", 16),
        new Person("Bob", 64)
    )

    val data = { people.map(p => p.toXml()) }

    def main(args : Array[String]){
        println(data)
    }
}

结果是:

Alice16Bob64

格式化结果(为了更好的阅读):

   
      Alice
      16
   
   
      Bob
      64
   

为完整起见,您还可以使用for..yield(或函数调用):

import scala.xml

case class Person(val name: String, val age: Int) {
  def toXml(): xml.Elem =
    { name }{ age }
}

def peopleToXml(people: List[Person]): xml.Elem = {
  {
    for {person <- people if person.age > 39}
      yield person.toXml
  }
}

val data = List(Person("joe",40),Person("mary", 35))
println(peopleToXml(data))

(Woody Folsom指出的固定误差)