I've written a trivial scala program to open an XML file.
Is there a way to get scala to validate the XML file against the schema file that it references? Currently my XML file doesn't follow the schema, so I'd expect to get errors on validation.
The XML file references the schema like this in the root element:
<items xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:noNamespaceSchemaLocation="items.xsd">
The scala code:
import scala.xml._
object HelloWorld {
def main(args: Array[String]) {
println("Hello, world! " + args.toList)
val start = System.currentTimeMillis
val data = XML.loadFile(args(0))
val stop = System.currentTimeMillis
Console.println("Took " + (stop-start)/1000.0 + "s to load " + args(0))
}
}
HelloWorld.main(args)