xml - 使用 javax.xml.xpath。使用 scala.xml

Question

是否可以将 javax.xml.xpath.XPathExpression 与 scala.xml.NodeSeq 一起使用？

我想要一个允许我表达如下内容的 API:

val xml = ...
val xpath = XPathFactory.newInstance.newXPath.compile(
    """/this/that/theOther[@abc="123"]""")
val selectedNodes: NodeSeq = xml.applyXpath(xpath)

Answer 1

Scala 采用函数式方法来搜索 XML。在相同的情况下，它不像 XPath 那样清晰，需要一些时间来适应它。例如:

scala> val myXml = <books><book category="1">first</book><book category="2">second</book><book category="1">third</book></books>
myXml: scala.xml.Elem = <books><book category="1">first</book><book category="2">second</book><book category="1">third</book></books>

scala> (myXml \ "book").filter { node => (node \\ "@category").text == "1" }
res24: scala.xml.NodeSeq = NodeSeq(<book category="1">first</book>, <book category="1">third</book>)