解析xml时如何在python中获取从根到特定元素的xpath

Question

我想列出 xml 中相对于它们的根的所有元素路径。例如

<A>
   <B>
     <C>Name</C>
     <D>Name</D>
   </B>
</A>

所以我想将它们列为：-

A/B/C
A/B/D

我能够使用 python 的“元素”对象解析 xml，但无法从中提取 xpath。有什么帮助吗？

Answer 1

可以构造解析树的父映射，然后用它构造所需的XPath：

import xml.etree.ElementTree as parser

def get_parent_map(root):
    return {c:p for p in root.iter() for c in p}

def extract_text_info(root, original_root):
    parent_map = get_parent_map(original_root)

    for child in root:
        if child.text is not None and len(child.text.strip()) > 0:
            c = child
            arr = []
            while c != original_root:
                arr.append(c.tag)
                c = parent_map[c]
            arr.append(original_root.tag)

            print('/'.join(arr[::-1]))
            print(child.text)

        extract_text_info(child, original_root)

那么我们有

xml = """<A>
       <B>
         <C>Name</C>
         <D>Name</D>
       </B>
     </A> """

root = parser.fromstring(xml)
extract_text_info(root, root)

> A/B/C
> Name
> A/B/D
> Name

Answer 2

sample.html

<A>
   <B>
     <C>Name1</C>
     <D>Name2</D>
   </B>
</A>

parse.py

from bs4 import BeautifulSoup

def get_root_elements(path_to_file):
    soup = BeautifulSoup(open(path_to_file), 'lxml')
    all_elements = soup.find_all()

    count_element_indices = [len(list(a.parents)) for a in all_elements]

    absolute_roots_index = min(
        (index for index, element in enumerate(count_element_indices)
            if element == max(count_element_indices)
        )
    )

    return all_elements[absolute_roots_index:]

def get_path(element):
    to_remove = ['[document]', 'body', 'html']
    path = [element.name] + [e.name for e in element.parents if e.name not in to_remove]

    return ' / '.join(path[::-1])

Python 外壳

In [1]: file = 'path/to/sample.html'

In [2]: run parse.py

In [3]: roots = get_root_elements(file)

In [4]: print(roots)
[<c>Name1</c>, <d>Name2</d>]

In [4]: for root in roots:
   ...:    print(get_path(root))
a / b / c
a / b / d

Answer 3

我想出的方法之一是通过代码。

 import xml.etree.ElementTree as ET


def parseXML(root,sm):
    sm = sm + "/" + root.tag[root.tag.rfind('}')+1:]
    for child in root:
      parseXML(child,sm)
    if len(list(root)) == 0:
      print(sm)

tree = ET.parse('test.xml')
root = tree.getroot()
parseXML(root,"")

不知道是否有相同的内置函数。