如何使用Scala验证XML文件的模式？

Question

我写了一个简单的scala程序来打开一个XML文件.

有没有办法让scala根据它引用的模式文件验证XML文件？目前我的XML文件不遵循架构,所以我希望在验证时出错.

XML文件在根元素中引用这样的模式:

<items xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:noNamespaceSchemaLocation="items.xsd">

scala代码:

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)

Answer 1

这是一篇博客文章,描述了如何在Scala中使用Java库进行模式验证:

http://sean8223.blogspot.com/2009/09/xsd-validation-in-scala.html

它归结为基本的重新实现XML.load:

import javax.xml.parsers.SAXParser
import javax.xml.parsers.SAXParserFactory
import javax.xml.validation.Schema
import javax.xml.validation.ValidatorHandler
import org.xml.sax.XMLReader

class SchemaAwareFactoryAdapter(schema:Schema) extends NoBindingFactoryAdapter {

  override def loadXML(source: InputSource): Elem = {
    // create parser
    val parser: SAXParser = try {
      val f = SAXParserFactory.newInstance()
      f.setNamespaceAware(true)
      f.setFeature("http://xml.org/sax/features/namespace-prefixes", true)
      f.newSAXParser()
    } catch {
      case e: Exception =>
        Console.err.println("error: Unable to instantiate parser")
        throw e
    }

    val xr = parser.getXMLReader()
    val vh = schema.newValidatorHandler()
    vh.setContentHandler(this)
    xr.setContentHandler(vh)

    // parse file
    scopeStack.push(TopScope)
    xr.parse(source)
    scopeStack.pop
    return rootElem.asInstanceOf[Elem]
  }
}