Scala:如何从某个集合创建XML节点

Question

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 = <people>{ /* what here?! */ }</people>

结果值应该是这样的:

<people><person><name>joe</name><age>40</age></person>
<person><name>mary</name><age>39</age></person></people>

Answer 1

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() = <person><name>{ name }</name><age>{ age }</age></person>
}

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

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

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

结果是:

<people><person><name>Alice</name><age>16</age></person><person><name>Bob</name><age>64</age></person></people>

格式化结果(为了更好的阅读):

<people>
   <person>
      <name>Alice</name>
      <age>16</age>
   </person>
   <person>
      <name>Bob</name>
      <age>64</age>
   </person>
</people>

Answer 2

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 =
    <person><name>{ name }</name><age>{ age }</age></person>
}

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

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

(Woody Folsom指出的固定误差)