如何更改Scala XML Element的属性

Question

我有一个XML文件,我想用脚本映射in的一些属性.例如:

<a>
  <b attr1 = "100" attr2 = "50"/>
</a>

可能有两个缩放的属性:

<a>
  <b attr1 = "200" attr2 = "100"/>
</a>

此页面提供了添加属性的建议,但没有详细说明使用函数映射当前属性的方法(这种方式会非常困难):http: //www.scalaclass.com/book/export/html/1

我想出的是手动创建XML(非scala)链表...类似于:

// a typical match case for running thru XML elements:
case  Elem(prefix, e, attributes, scope, children @ _*) => {
 var newAttribs = attributes
 for(attr <- newAttribs)  attr.key match {
  case "attr1" => newAttribs = attribs.append(new UnprefixedAttribute("attr1", (attr.value.head.text.toFloat * 2.0f).toString, attr.next))
  case "attr2" => newAttribs = attribs.append(new UnprefixedAttribute("attr2", (attr.value.head.text.toFloat * 2.0f).toString, attr.next))
  case _ =>
 }
 Elem(prefix, e, newAttribs, scope, updateSubNode(children) : _*)  // set new attribs and process the child elements
}

它的丑陋,冗长,不必要地重新排序输出中的属性,由于一些糟糕的客户端代码,这对我当前的项目是不利的.有一种scala式的方式吗？

Answer 1

好的,尽力而为,Scala 2.8.我们需要重建属性,这意味着我们必须正确地分解它们.让我们为此创建一个函数:

import scala.xml._

case class GenAttr(pre: Option[String], 
                   key: String, 
                   value: Seq[Node], 
                   next: MetaData) {
  def toMetaData = Attribute(pre, key, value, next)
}

def decomposeMetaData(m: MetaData): Option[GenAttr] = m match {
  case Null => None
  case PrefixedAttribute(pre, key, value, next) => 
    Some(GenAttr(Some(pre), key, value, next))
  case UnprefixedAttribute(key, value, next) => 
    Some(GenAttr(None, key, value, next))
}

接下来,让我们将链接的属性分解为一个序列:

def unchainMetaData(m: MetaData): Iterable[GenAttr] = 
  m flatMap (decomposeMetaData)

此时,我们可以轻松操作此列表:

def doubleValues(l: Iterable[GenAttr]) = l map {
  case g @ GenAttr(_, _, Text(v), _) if v matches "\\d+" => 
    g.copy(value = Text(v.toInt * 2 toString))
  case other => other
}

现在,再次链接它:

def chainMetaData(l: Iterable[GenAttr]): MetaData = l match {
  case Nil => Null
  case head :: tail => head.copy(next = chainMetaData(tail)).toMetaData
}

现在,我们只需要创建一个函数来处理这些事情:

def mapMetaData(m: MetaData)(f: GenAttr => GenAttr): MetaData = 
  chainMetaData(unchainMetaData(m).map(f))

所以我们可以像这样使用它:

import scala.xml.transform._

val attribs = Set("attr1", "attr2")
val rr = new RewriteRule {
  override def transform(n: Node): Seq[Node] = (n match {
    case e: Elem =>
      e.copy(attributes = mapMetaData(e.attributes) {
        case g @ GenAttr(_, key, Text(v), _) if attribs contains key =>
          g.copy(value = Text(v.toInt * 2 toString))
        case other => other
      })
    case other => other
  }).toSeq
}
val rt = new RuleTransformer(rr)

最后让你做你想要的翻译:

rt.transform(<a><b attr1="100" attr2="50"></b></a>)

如果符合以下条件,所有这些都可以

• 属性实际上定义了前缀,键和值,带有可选前缀
• 属性是一个序列,而不是一个链
• 属性有map,mapKeys,mapValues
• Elem有一个mapAttribute

Answer 2

这是使用Scala 2.10的方法:

import scala.xml._
import scala.xml.transform._

val xml1 = <a><b attr1="100" attr2="50"></b></a>

val rule1 = new RewriteRule {
  override def transform(n: Node) = n match {
    case e @ <b>{_*}</b> => e.asInstanceOf[Elem] % 
      Attribute(null, "attr1", "200", 
      Attribute(null, "attr2", "100", Null))
    case _ => n 
  }
}

val xml2 = new RuleTransformer(rule1).transform(xml1)

Answer 3

所以,如果我在你的位置,我认为我真正想要写的是:

case elem: Elem => elem.copy(attributes=
  for (attr <- elem.attributes) yield attr match {
    case attr@Attribute("attr1", _, _) =>
      attr.copy(value=attr.value.text.toInt * 2)
    case attr@Attribute("attr2", _, _) =>
      attr.copy(value=attr.value.text.toInt * -1)
    case other => other
  }
)

这有两个原因:开箱即用:

1. Attribute没有一个有用的copy方法,和
2. 映射到一个MetaData产生一个Iterable[MetaData]而不是一个MetaData简单的事情elem.copy(attributes=elem.attributes.map(x => x))就会失败.

要解决第一个问题,我们将使用隐式方法将更好的复制方法添加到Attribute:

implicit def addGoodCopyToAttribute(attr: Attribute) = new {
  def goodcopy(key: String = attr.key, value: Any = attr.value): Attribute =
    Attribute(attr.pre, key, Text(value.toString), attr.next)
}

copy由于具有该名称的方法已经存在,因此无法命名,因此我们只需调用它即可goodcopy.(另外,如果您创建的值Seq[Node]不是应该转换为字符串的值,那么您可能会更加小心value,但对于我们当前的目的,它不是必需的.)

为了解决第二个问题,我们将使用一个隐式来解释如何创建一个MetaData来自Iterable[MetaData]:

implicit def iterableToMetaData(items: Iterable[MetaData]): MetaData = {
  items match {
    case Nil => Null
    case head :: tail => head.copy(next=iterableToMetaData(tail))
  }
}

然后你可以编写代码,就像我在开头提出的那样:

scala> val elem = <b attr1 = "100" attr2 = "50"/>
elem: scala.xml.Elem = <b attr1="100" attr2="50"></b>

scala> elem.copy(attributes=
     |   for (attr <- elem.attributes) yield attr match {
     |     case attr@Attribute("attr1", _, _) =>
     |       attr.goodcopy(value=attr.value.text.toInt * 2)
     |     case attr@Attribute("attr2", _, _) =>
     |       attr.goodcopy(value=attr.value.text.toInt * -1)
     |     case other => other
     |   }
     | )
res1: scala.xml.Elem = <b attr1="200" attr2="-50"></b>