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,因此我希望验证时会出错。
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)
答案 0 :(得分:6)
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
归结为XML.load
的基本重新实现:
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]
}
}
答案 1 :(得分:2)
我认为你不能用Scala库做到这一点。 但你绝对可以使用Java库。只需谷歌“java架构验证”,你会发现很多选项
答案 2 :(得分:2)
以下是对2.8.0(或2.8.1)中的次要API更改的修改:
import org.xml.sax.InputSource
import scala.xml.parsing.NoBindingFactoryAdapter
import scala.xml.{TopScope, Elem}
import javax.xml.parsers.{SAXParserFactory, SAXParser}
import javax.xml.validation.Schema
class SchemaAwareFactoryAdapter(schema: Schema) extends NoBindingFactoryAdapter {
override def loadXML(source: InputSource, parser: SAXParser) = {
val reader = parser.getXMLReader()
val handler = schema.newValidatorHandler()
handler.setContentHandler(this)
reader.setContentHandler(handler)
scopeStack.push(TopScope)
reader.parse(source)
scopeStack.pop
rootElem.asInstanceOf[Elem]
}
override def parser: SAXParser = {
val factory = SAXParserFactory.newInstance()
factory.setNamespaceAware(true)
factory.setFeature("http://xml.org/sax/features/namespace-prefixes", true)
factory.newSAXParser()
}
}
申请也略有不同:
val factory = SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI)
val xsdStream = getClass.getResourceAsStream("/foo.xsd")
val schema = factory.newSchema(new StreamSource(stream))
val source = getClass.getResourceAsStream("baz.xml")
val xml = new SchemaAwareFactoryAdapter(schema).load(source)