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如何将scala.xml.Elem转换为与javax.xml API兼容的东西?

interop scala xml java

6

overthink · 技术社区 · 16 年前

scala.xml.Elem ),我想将它与一些标准的JavaXMLAPI(特别是 SchemaFactory ). 看起来像是把我的 Elem javax.xml.transform.Source 这是我需要做的,但我不确定。我可以找到各种方法来有效地写出我的并将其读入与Java兼容的内容,但我想知道是否有一种更优雅(希望更高效)的方法?

Scala代码:

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)))

1 回复 | 直到 16 年前

1

2

Steven Merrill 16 年前

你想要的是 -您只需轻轻地告诉Scala编译器如何从 scala.xml.Elem 到 javax.xml.transform.stream.StreamSource .

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)

它的效率并没有提高,但一旦你得到了隐式方法定义,它肯定会更漂亮。