scala - XML 插入/更新 [英] scala - XML insert/update

问题描述

你知道根据 XPath 插入和(或)更新节点的任何 Scala API 吗?例如，对于给定的节点和 XPath，此 API 将使用新节点创建 XML 副本

do you know any Scala API to insert and (or) update Nodes according to XPath? e.g for a given Node and XPath, this API would create a copy of XML with new node

谢谢

推荐答案

你可以使用 RewriteRule 来做到这一点，2.10.3 文档.

You can use the RewriteRule to do this, 2.10.3 documentation.

val cats = <Cats>
  <Cat Name="Floyd"/>
  <Cat Name="Onyx"/>
</Cats>

然后假设 RewriteRule

class AddCat(name: String) extends RewriteRule {
   override def transform(n: Node): Seq[Node] = n match {
     case e: Elem if e.label == "Cats" =>
       val cats = (e \ "Cat")
       val newCat = <Cat Name={name}/>
       new Elem(e.prefix, "Cats", e.attributes, e.scope, e.minimizeEmpty, (cats ++ newCat).toSeq:_*)
     case x => x
   }
 }

那你就可以了，

val rule = new RuleTransformer(new AddCat("Stevie"))
rule.transform(cats)
res2: Seq[scala.xml.Node] = List(<Cats><Cat Name="Floyd"/><Cat Name="Onyx"/><Cat Name="Stevie"/></Cats>)

如果你想改变一个属性也一样

Similarly if you wanted to change an attribute

class AddLastName(name: String, lastName: String) extends RewriteRule {
  override def transform(n: Node): Seq[Node] = n match {
    case e: Elem if e.label == "Cat" && (e \ "@Name" text).equals(name) =>
      val cat: String = e.attributes("Name").head.text
      e % Attribute(None, "Name", Text(s"$name $lastName"), Null)
    case x => x
  }
}

val rule = new RuleTransformer(new AddLastName("Stevie", "Nicks"))
rule.transform(cats)
res3: Seq[scala.xml.Node] = List(<Cats><Cat Name="Floyd"/><Cat Name="Onyx"/><Cat Name="Stevie Nicks"/></Cats>)

这两种方法都可以满足您的需求.困难的部分是弄清楚如何获取子节点，然后构建父节点.

Both of these approaches would do what you're looking for. The hard part is figuring out how to get at the children, then build the parent node.

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