scala:如何从某个集合创建XML节点

因为它是一个函数式编程语言数组。map可能是您要查找的:

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:

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

格式化结果(为了更好地阅读):

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

为了完整性,还可以使用..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记录的修正错误)

实际上,yield person.toxml()一行不是为我编译的,而是yield person.toxml(不带括号)编译的。原始版本抱怨“重载方法值应用”,即使我更改“to xml”的定义以显式返回scala.xml.elem