如何判断一个Nokogiri::XML::Node是否包含另一个的部分节点？

我使用Ruby的Nokogiri库，自定义了ConstituencyXMLNode类继承自Nokogiri::XML::Node，现在有两个节点：

节点结构示例 #

pp_leaf 节点 #

#(ConstituencyXMLNode:0xc3c8 {
  name = "PP",
  children = [
    #(ConstituencyXMLNode:0xc3dc { name = "IN", children = [ #(Text "out")] }),
    #(ConstituencyXMLNode:0xc3f0 {
      name = "PP",
      children = [
        #(ConstituencyXMLNode:0xc404 { name = "IN", children = [ #(Text "of")] }),
        #(ConstituencyXMLNode:0xc418 {
          name = "NP",
          children = [
            #(ConstituencyXMLNode:0xc42c { name = "JJ", children = [ #(Text "non-living")] }),
            #(ConstituencyXMLNode:0xc440 { name = "NNS", children = [ #(Text "resources")] })]
          })]
      })]
  })

current_leaf 节点 #

#(ConstituencyXMLNode:0xc3f0 {
  name = "PP",
  children = [
    #(ConstituencyXMLNode:0xc404 { name = "IN", children = [ #(Text "of")] }),
    #(ConstituencyXMLNode:0xc418 {
      name = "NP",
      children = [
        #(ConstituencyXMLNode:0xc42c { name = "JJ", children = [ #(Text "non-living")] }),
        #(ConstituencyXMLNode:0xc440 { name = "NNS", children = [ #(Text "resources")] })]
      })]
  })

我需要判断pp_leaf是否包含current_leaf对应的节点结构。注意到Nokogiri::XML::NodeSet有include?方法，但不清楚如何将单个Node转为NodeSet，且Node本身没有直接判断结构包含的操作符，求解决思路。

解决思路 #

1. 区分对象引用匹配与结构匹配 #

如果是判断current_leaf是否是pp_leaf的后代节点（同一对象引用），直接用NodeSet的include?即可：

# 检查current_leaf是否是pp_leaf的后代节点
pp_leaf.descendants.include?(current_leaf)

如果要判断的是结构完全一致的节点（即使是不同对象），则需要递归比较节点结构。

2. 递归比较节点结构 #

编写辅助方法递归校验节点的名称、子节点数量及内容：

def structure_matches?(node_a, node_b)
  # 节点名称必须一致
  return false unless node_a.name == node_b.name
  # 子节点数量必须一致
  return false unless node_a.children.size == node_b.children.size

  # 逐个比较子节点
  node_a.children.zip(node_b.children).all? do |child_a, child_b|
    if child_a.text? && child_b.text?
      # 文本节点忽略多余空格后比较
      child_a.text.strip == child_b.text.strip
    else
      # 非文本节点递归比较结构
      structure_matches?(child_a, child_b)
    end
  end
end

# 检查pp_leaf的后代中是否有结构匹配current_leaf的节点
contains_target = pp_leaf.descendants.any? { |node| structure_matches?(node, current_leaf) }

3. 快速序列化比较（简单场景） #

如果节点结构简单、无复杂属性，可将节点序列化为XML字符串后判断包含关系：

# 序列化时去除多余空格，避免格式差异影响判断
current_xml = current_leaf.to_xml.strip.gsub(/\s+/, ' ')
pp_xml = pp_leaf.to_xml.strip.gsub(/\s+/, ' ')

contains_target = pp_xml.include?(current_xml)

4. 将单个Node转为NodeSet #

如果确实需要将单个Node转为NodeSet，直接通过构造方法创建即可：

current_node_set = Nokogiri::XML::NodeSet.new(current_leaf.document, [current_leaf])

但注意NodeSet#include?仅判断对象引用是否存在，不校验结构一致性。

内容的提问来源于stack exchange，提问作者Kelsey Hannan