Scala：將org.w3c.dom.Document轉換爲scala.xml.NodeSeq

Question

標題非常明瞭。如何將org.w3c.dom.Document的實例轉換爲Scala NodeSeq，以享受它的便利？

乾杯
帕爾薩

Answer 1

def asXml(dom: org.w3c.dom.Node): Node = { 
    val dom2sax = new DOM2SAX(dom) 
    val adapter = new NoBindingFactoryAdapter 
    dom2sax.setContentHandler(adapter) 
    dom2sax.parse() 
    return adapter.rootElem 
    } 

Answer 2

我寫了這個代碼前陣子在另一個方向走，從Scala的節點到節點DOM4J。這表明在遞歸在樹的基本思路，應該是很容易適應：

implicit def scalaToDom4j(n : Node) : DElem = { 

    def inner(n : Node) : Option[DNode] = { 
    n match { 
     case e : Elem => 
     val elem = DocumentHelper.createElement(e.label) 
     for(c <- e.child) yield inner(c) collect { 
      case Some(child) => elem.add(child) 
     } 
     Some(elem) 
     //as Scala's xml is type-aware, text might not actually be a Text node, 
     //but an Atom of some other type 
     case t : Atom[_] => 
     Some(DocumentHelper.createText(t.data.toString)) 
     case x => None 
    } 
    } 

    //Attempt the conversion. Throw an exception if something has gone badly wrong 
    //inner returns an Option[DNode], but the expected top-level type is a DElem 
    // (which is a subclass of DNode) 
    //so we also validate this. 
    inner(trim(n)) map (_.asInstanceOf[DElem]) getOrElse (error("xml failed")) 
} 

Answer 3

我有遞歸模式匹配代碼類同的scalaToDom4j（n）的上述功能，只有在薩克森XdmNode到Scala節點方向： https://github.com/LeifW/MusicPath/blob/master/src/main/scala/org/musicpath/saxon2scala/Saxon2Scala.scala

目前它只是使文本，元素和非名稱空間屬性節點，但它應該很容易推廣/完成。

Answer 4

IttayD的答案適用於所有w3c XML - 除了dom4j w3c兼容xmls。所有W3C類型的以下作品：

def asXml(dom: _root_.org.w3c.dom.Node): Node = { 
    val source = new DOMSource(dom) 
    val adapter = new NoBindingFactoryAdapter 
    val saxResult = new SAXResult(adapter) 
    val transformerFactory = javax.xml.transform.TransformerFactory.newInstance() 
    val transformer = transformerFactory.newTransformer() 
    transformer.transform(source, saxResult) 
    adapter.rootElem 
    }