Question

def loadXml(node: Node): Unit = {

  val children = node.child

  children.foreach(child => {
    var tag = child.label

    //if owner tag, load the owen
    if (tag == "zip")
    {
      loadZipXML(child)
    }
    else if (tag == "owner")
    {
      //if owner tag, make a new pet and have it load the info it wants, then add it to the list
      val owner = Owner()
      owner.loadXml(child)
      insurance += owner
    }
  })
}

我有以下代码，并且正在使用此XML：

<?xml version='1.0' encoding='UTF-8'?>
<insurance>
    <zip code="57701">
        <owner name="Harold">
        </owner>
        <owner name="Bob">
        </owner>
        <owner name="Indiana Jones">
        </owner>
        <owner name="Darth Vader">
        </owner>
    </zip>
    <zip code="57702">
        <owner name="Sue">
        </owner>
        <owner name="Captain Kirk">
        </owner>
    </zip>
    <zip code="57703">
    </zip>
</insurance>

我可以拉邮政编码。但是，每当我得到所有者时，标签就会变成#PCDATA。我知道这意味着它是一个拥有更多数据的孩子，但是我该如何抓住该标签，然后继续遍历XML文件？

Answer 1

不确定为什么需要XML标签，但这听起来像是您试图将数据编组到“所有者”数据结构中。例如，考虑以下“所有者”类：

final case class Owner(
  label: String, 
  name:  String, 
  text:  String, 
  zip:   String)

我可能会通过以下方式遍历XML：

val insurance = scala.xml.XML.load("insurance.xml")
val owners = 
  for {
    zip   <- insurance \ "zip"
    owner <- zip \ "owner"
  } yield {
    Owner(
      label = owner.label, 
      name  = owner \@ "name",
      text  = owner.text.trim,
      zip   = zip \@ "code"
    )
  }

打印文件：

owners.foreach(println)

输出：

    Owner(owner,Harold,,57701)
    Owner(owner,Bob,,57701)
    Owner(owner,Indiana Jones,,57701)
    Owner(owner,Darth Vader,,57701)
    Owner(owner,Sue,,57702)
    Owner(owner,Captain Kirk,,57702)

如何从Scala中的“ #PCDATA”获取数据？

1 个答案: