How do I create an XML root node in Scala without a literal element name?

Question

I'm looking to create a document like this:

<root/>

That I can add children to programatically. Theoretically, it would look like this:

val root_node_name = "root"
val doc = <{root_node_name}/>

But that doesn't seem to work:

error: not found: value <

So, what I tried instead was this:

val root_node_name = "root"
val doc = new scala.xml.Elem(null, root_node_name, null, scala.xml.TopScope, null)  

That compiles but at runtime I get this null pointer exception:

java.lang.NullPointerException
at scala.xml.Utility$.toXML(Utility.scala:201)
at scala.xml.Utility$$anonfun$sequenceToXML$2.apply(Utility.scala:235)
at scala.xml.Utility$$anonfun$sequenceToXML$2.apply(Utility.scala:235)
at scala.Iterator$class.foreach(Iterator.scala:414)
at scala.runtime.BoxedArray$AnyIterator.foreach(BoxedArray.scala:45)
at scala.Iterable$class.foreach(Iterable...

I'm using Scala 2.8. Any examples of how to pull this off? Thanks.

Answer 1

You should pass the empty list for attributes ( scala.xml.Null ) and if you don't want any children, you shouldn't even include the final argument. You want an empty list of children, not a single child that happens to be null . So:

scala> val root_node_name = "root"
root_node_name: java.lang.String = root

scala> val doc = new scala.xml.Elem(null, root_node_name, scala.xml.Null , scala.xml.TopScope)
doc: scala.xml.Elem = <root></root>

Answer 2

On 2.8 you can do this:

scala> val r = <root/>
r: scala.xml.Elem = <root></root>

scala> r.copy(label="bar")
res0: scala.xml.Elem = <bar></bar>

So if your initial document is <root/> , then just use a literal. If you need to be able to set the label at runtime, you can define a method like this:

def newRoot(label:String) = {val r = <root/>; r.copy(label=label) }