JSON to XML in Scala and dealing with Option() result

Question

Consider the following from the Scala interpreter:

scala> JSON.parseFull(\"\"\"{\"name\":\"jack\",\"greeting\":\"hello world\"}\"\"\")
res6: Option[Any] = Som         


        

Answer 1

An Option is returned because parseFull has different possible return values depending on the input, or it may fail to parse the input at all (giving None). So, aside from an optional Map which associates keys with values, an optional List can be returned as well if the JSON string denoted an array.

Example:

scala> import scala.util.parsing.json.JSON._
import scala.util.parsing.json.JSON._

scala> parseFull("""{"name":"jack"}""")
res4: Option[Any] = Some(Map(name -> jack))

scala> parseFull("""[ 100, 200, 300 ]""")
res6: Option[Any] = Some(List(100.0, 200.0, 300.0))

You might need pattern matching in order to achieve what you want, like so:

scala> parseFull("""{"name":"jack","greeting":"hello world"}""") match {
     |   case Some(m) => Console println ("Got a map: " + m)
     |   case _ =>
     | }
Got a map: Map(name -> jack, greeting -> hello world)

Now, if you want to generate XML output, you can use the above to iterate over the key/value pairs:

import scala.xml.XML

parseFull("""{"name":"jack","greeting":"hello world"}""") match {
  case Some(m: Map[_,_]) =>
    
      {
        m map { case (k,v) =>
          XML.loadString("<%s>%s".format(k,v,k))
        }
      }
    

  case _ =>
}