使用 ElementTree 查找节点的路径答案

【问题标题】：Find path to the node using ElementTree使用 ElementTree 查找节点的路径
【发布时间】：2020-03-26 08:30:00
【问题描述】：

使用ElementTree，我可以打印特定标签的每次出现（在我的情况下为ExpertSettingsSg ):

#!/usr/bin/env python3

import xml.etree.ElementTree as ET

root = ET.parse('mydoc.xml').getroot()

for children in root:
    value=children.findall('.//ExpertSettingsSg')#tag I'm looking for
    for settings in value:
        if settings.text is not None:
            print(settings.text)

但我没有找到打印发生路径的方法。因为我的 XML 文件有很多级别，并且因为ExpertSettingsSg 几乎可以在每个级别，所以我需要知道ExpertSettingsSg 的来源。我正在寻找类似的东西

配置路径 xxxxxx = /root/xxx/aaaa/bbbb

如果ElementTree 无法实现，还有其他库可以解决问题吗？

谢谢

【问题讨论】：

标签： python-3.x xml elementtree

【解决方案1】：

如果您已经有了节点，您可以遍历树并收集路径（借用 @valdi-bo 的示例）：

from xml.etree import ElementTree as ET

txt ='''<main>
  <x>
    <a>
      <ExpertSettingsSg id="1">x1</ExpertSettingsSg>
    </a>
    <b>
      <dummy>xxxx</dummy>
    </b>
  </x>
  <y>
    <c>
      <dummy>xxxx</dummy>
    </c>
    <d>
      <ExpertSettingsSg id="2">x2</ExpertSettingsSg>
    </d>
    <e>
      <ExpertSettingsSg id="3"/>
    </e>
  </y>
</main>'''


def node_walk(root: ET.Element):
    path_to_node = []
    node_stack = [root]
    while node_stack:
        node = node_stack[-1]
        if path_to_node and node is path_to_node[-1]:
            path_to_node.pop()
            node_stack.pop()
            yield (path_to_node, node)
        else:
            path_to_node.append(node)
            for child in reversed(node):
                node_stack.append(child)


root = ET.ElementTree(ET.fromstring(txt))

for node in root.findall('.//ExpertSettingsSg'):
    for node_path, n in node_walk(root.getroot()):
        if n is node:
            xpath = "/".join(["."] + [n.tag for n in node_path[1:]] + [n.tag])
            print(xpath, node)
            # NOTE: Assert is to just show that the xpath is correct.
            assert root.getroot().find(xpath) == node

你会得到这样的输出：

./x/a/ExpertSettingsSg <Element 'ExpertSettingsSg' at 0x102cf5b80>
./y/d/ExpertSettingsSg <Element 'ExpertSettingsSg' at 0x102cf5db0>
./y/e/ExpertSettingsSg <Element 'ExpertSettingsSg' at 0x102cf5e50>

我们可以步行一次并收集所有具有路径的相关节点，而不是多次步行，如下所示：

xpaths = []
for node_path, n in node_walk(root.getroot()):
    if n.tag == "ExpertSettingsSg":
        xpath = "/".join(["."] + [n.tag for n in node_path[1:]] + [n.tag])
        xpaths.append(xpath)

for xpath in xpaths: 
    node = root.getroot().find(xpath)
    print(xpath, node)

【讨论】：