Question

好吧，这对我（Scala的新秀）以及我的同事（在Scala更先进）感到困惑。 Scala 2.8.0。以下是该问题的演示：

// I've got a var with some XML in it
scala> qq2
res9: scala.xml.Elem = <a><a1>A1</a1><bs><b>B1</b><c>C1</c><d>D1</d></bs></a>

// I can extract sub-elements
scala> (qq2 \ "bs")
res10: scala.xml.NodeSeq = NodeSeq(<bs><b>B1</b><c>C1</c><d>D1</d></bs>)

// but if I try to match against this NodeSeq, it fails to match
scala> (qq2 \ "bs") match {case <bs>{x @ _*}</bs> => 
            for (xx <- x) println("matched " + xx) }      
scala.MatchError: <bs><b>B1</b><c>C1</c><d>D1</d></bs>
        at .<init>(<console>:7)
        at ...

// but if I just type in the XML directly, it works as expected
scala> <bs><b>B1</b><c>C1</c><d>D1</d></bs> match {
          case <bs>{x @ _*}</bs> => for (xx <- x) println("matched " + xx) }
matched <b>B1</b>
matched <c>C1</c>
matched <d>D1</d>

// presumably because it's of type Elem, not NodeSeq
scala> <bs><b>B1</b><c>C1</c><d>D1</d></bs>
res13: scala.xml.Elem = <bs><b>B1</b><c>C1</c><d>D1</d></bs>

所以，两个问题。一：wtf？为什么会这样？二：我似乎无法找到将NodeSeq转换为Elem的方法，因此匹配将起作用。什么是正确的方法？

Answer 1

NodeSeq是Node的集合，而不是单个节点：

scala> (<a><b>1</b><b>2</b></a>) \ "b"
res0: scala.xml.NodeSeq = NodeSeq(<b>1</b>, <b>2</b>)

所以你必须匹配节点：

scala> ((<a><b>1</b><b>2</b></a>) \ "b").map(_ match {
     |   case <b>{x}</b> => true
     |   case _ => false
     | })
res24: scala.collection.immutable.Seq[Boolean] = List(true, true)

（节点往往是Elems，所以这很好。我不知道分离背后的原因;我猜一些节点与它们的关联性比Elem少。）

Answer 2

方法\返回一系列有效答案，而不是单个元素。这里：

scala> val qq2 = <a><a1>A1</a1><bs><b>B1</b><c>C1</c><d>D1</d></bs></a>
qq2: scala.xml.Elem = <a><a1>A1</a1><bs><b>B1</b><c>C1</c><d>D1</d></bs></a>

scala> (qq2 \ "bs") match {case Seq(<bs>{x @ _*}</bs>) => //<-I added a Seq()
     |     for (xx <- x) println("matched " + xx) }
matched <b>B1</b>
matched <c>C1</c>
matched <d>D1</d>

NodeSeq匹配失败，但等效的Elem匹配成功 - 为什么？怎么修？

2 个答案: