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NodeSeq匹配失败,但等效元素匹配成功-为什么?如何修复?

match scala xml

3

Harlan · 技术社区 · 15 年前

好吧,这让我(Scala的新手)和我的同事(Scala的更高级)都很难堪。Scala 2.8.0版。下面是问题的演示:

// 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,有两个问题。一:世贸论坛?为什么是这样?第二:我似乎找不到一种方法把NodeSeq转换成Elem,这样匹配就可以了。正确的方法是什么?

2 回复 | 直到 15 年前

1

4

Ken Bloom 15 年前

方法 \ 返回有效答案的序列,而不是单个元素。在这里:

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>

2

6

Rex Kerr 15 年前

A NodeSeq 是一个 Node s、没有一个节点:

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

因此必须在节点上匹配:

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)