Scala: how to create XML nodes from some collection

Question

If you have something like:

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

How do you create an XML value with that data as

Answer 1

For completeness, you can also use for..yield (or function calls):

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))

(fixed error noted by Woody Folsom)

Answer 2

As it's a functional programming language Array.map is probably what you're looking for:

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)
    }
}

Results in:

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

A formatted result (for a better read):

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

Answer 3

Actually, the line yield person.toXml() does not compile for me, but yield person.toXml (without the parentheses) does. The original version complains of 'overloaded method value apply' even if I change the def of 'ToXml' to explicitly return a scala.xml.Elem