Question

I have a Scala representation of some XML (i.e. a scala.xml.Elem), and I’d like to use it with some of the standard Java XML APIs (specifically SchemaFactory). It looks like converting my Elem to a javax.xml.transform.Source is what I need to do, but I’m not sure. I can see various ways to effectively write out my Elem and read it into something compatible with Java, but I’m wondering if there’s a more elegant (and hopefully more efficient) approach?

Scala code:

import java.io.StringReader
import javax.xml.transform.stream.StreamSource
import javax.xml.validation.{Schema, SchemaFactory}
import javax.xml.XMLConstants

val schemaXml = <xsd:schema xmlns:xsd="http://www.w3.org/2001/XMLSchema">
                  <xsd:element name="foo"/>
                </xsd:schema>
val schemaFactory = SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);

// not possible, but what I want:
// val schema = schemaFactory.newSchema(schemaXml)

// what I'm actually doing at present (ugly)
val schema = schemaFactory.newSchema(new StreamSource(new StringReader(schemaXml.toString)))

Answer 1

What you want is possible – you just have to gently tell the Scala compiler how to go from scala.xml.Elem to javax.xml.transform.stream.StreamSource by declaring an implicit method.

import java.io.StringReader
import javax.xml.transform.stream.StreamSource
import javax.xml.validation.{Schema, SchemaFactory}
import javax.xml.XMLConstants
import scala.xml.Elem

val schemaXml = <xsd:schema xmlns:xsd="http://www.w3.org/2001/XMLSchema">
                  <xsd:element name="foo"/>
                </xsd:schema>
val schemaFactory = SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);

implicit def toStreamSource(x:Elem) = new StreamSource(new StringReader(x.toString))

// Very possible, possibly still not any good:
val schema = schemaFactory.newSchema(schemaXml)

It isn’t any more efficient, but it sure is prettier once you get the implicit method definition out of the way.