如何更改 Scala XML 元素上的属性 [英] How to change attribute on Scala XML Element

问题描述

我有一个 XML 文件，我想用脚本映射其中的一些属性.例如:

I have an XML file that I would like to map some attributes of in with a script. For example:

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

可能具有按 2 倍缩放的属性:

might have attributes scaled by a factor of two:

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

这个页面有一个关于添加属性的建议，但没有详细说明用函数映射当前属性的方法(这种方式会变得非常困难):http://www.scalaclass.com/book/export/html/1

This page has a suggestion for adding attributes but doesn't detail a way to map a current attribute with a function (this way would make that very hard): http://www.scalaclass.com/book/export/html/1

我想出的是手动创建 XML(非 scala)链接列表......类似于:

What I've come up with is to manually create the XML (non-scala) linked-list... something like:

// 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
}

它可怕、冗长且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，这对我当前的项目不利.有没有一种 Scala 式的方法来做到这一点?

Its hideous, wordy, and needlessly re-orders the attributes in the output, which is bad for my current project due to some bad client code. Is there a scala-esque way to do this?

推荐答案

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分解它们.让我们为此创建一个函数:

Ok, best effort, Scala 2.8. We need to reconstruct attributes, which means we have to decompose them correctly. Let's create a function for that:

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:

Next, let's decompose the chained attributes into a sequence:

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

此时，我们可以轻松操作这个列表:

At this point, we can easily manipulate this list:

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:

Now, we only have to create a function to take care of these things:

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

所以我们可以这样使用它:

So we can use it like this:

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)

最终让你做你想要的翻译:

Which finally let you do the translation you wanted:

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

如果满足以下条件，所有这些都可以简化:

All of this could be simplified if:

• 属性实际定义了前缀、键和值，带有可选的前缀
• 属性是一个序列，而不是一个链
• 属性有一个地图、mapKeys、mapValues
• Elem 有一个 mapAttribute

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