在 Scala 中查找名称中带有通配符的 XML 节点 [英] Find XML nodes with wild card in name in Scala

问题描述

我有以下 XML:

<data>
    <a>...</a>
    <b1>...</b1>
    <c>...</c>
    <b2>...</b2>
    <d>...</d>
    <b3>...</b2>
</data>

在 Scala 中，如何从 data 节点(即 Elem 对象)中提取以字符串b"开头的节点?在这种情况下，所需的值是三个节点的序列:

In Scala, how do I extract the nodes that start with the string "b" from the data node (i.e., Elem object)? In this case, the desired value is a sequence of three nodes:

[<b1>...</b1>, <b2>...</b2>, <b3>...</b3]

我已经试过了，但它不能编译:

I've tried this, but it doesn't compile:

val orderNodes: NodeSeq = /data/*[starts-with(name(), "b")]

推荐答案

小文档最简单的解决方案是用你需要的谓词过滤节点序列:

The simplest solution for small documents is to filter the sequence of nodes with the predicate you need:

val data = <data>
  <a>...</a>
  <b1>...</b1>
  <c>...</c>
  <b2>...</b2>
  <d>...</d>
  <b3>...</b3>
</data>

scala> (data \ "_").filter(_.label.startsWith("b"))
res1: scala.xml.NodeSeq = NodeSeq(<b1>...</b1>, <b2>...</b2>, <b3>...</b3>)

elem \ label 语法返回一个节点序列，其名称与 label 完全相同.而 elem \ "_" 是该语法的一个特例，它返回所有子元素.然后，您可以像处理任何普通的 Scala 集合一样处理该节点序列.

elem \ label syntax returns a sequence of nodes, that have the name exactly equal to label. And elem \ "_" is a special case of that syntax, that returns all the child elements. Then you can work with that sequence of nodes like with any normal Scala collection.

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