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ergoithz/xpath_soup.py

Last active January 11, 2025 16:34
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Generate unique XPATH for BeautifulSoup element
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xpath_soup.py
#!/usr/bin/python
# -*- coding: utf-8 -*-
import bs4
def xpath_soup(element):
# type: (typing.Union[bs4.element.Tag, bs4.element.NavigableString]) -> str
"""
Generate xpath from BeautifulSoup4 element.
:param element: BeautifulSoup4 element.
:type element: bs4.element.Tag or bs4.element.NavigableString
:return: xpath as string
:rtype: str
Usage
-----
>>> import bs4
>>> html = (
... '<html><head><title>title</title></head>'
... '<body><p>p <i>1</i></p><p>p <i>2</i></p></body></html>'
... )
>>> soup = bs4.BeautifulSoup(html, 'html.parser')
>>> xpath_soup(soup.html.body.p.i)
'/html/body/p[1]/i'
>>> import bs4
>>> xml = (
... '<?xml version="1.0" encoding="UTF-8"?>'
... '<doc xmlns:ns1="http://localhost/ns1"'
... ' xmlns:ns2="http://localhost/ns2">'
... '<ns1:elm/><ns2:elm/><ns2:elm/></doc>'
... )
>>> soup = bs4.BeautifulSoup(xml, 'lxml-xml')
>>> xpath_soup(soup.doc.find('ns2:elm').next_sibling)
'/doc/ns2:elm[2]'
"""
components = []
target = element if element.name else element.parent
for node in (target, *target.parents)[-2::-1]: # type: bs4.element.Tag
tag = '%s:%s' % (node.prefix, node.name) if node.prefix else node.name
siblings = node.parent.find_all(tag, recursive=False)
components.append(tag if len(siblings) == 1 else '%s[%d]' % (tag, next(
index
for index, sibling in enumerate(siblings, 1)
if sibling is node
)))
return '/%s' % '/'.join(components)
if __name__ == '__main__':
import doctest
doctest.testmod(verbose=True, raise_on_error=True)
@funway
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funway commented Dec 28, 2023

Hi @ergoithz,
I faced a little problem if the XML element had a namespace prefix, the xpath returned by this function would not include the prefix. Looks like that:

<?xml version="1.0" encoding="UTF-8"?>
<xfa:datasets xmlns:xfa="http://www.xfa.org/schema/xfa-data/1.0/">
    <data>ABC</data>
</xfa:datasets>

The output of tag will be /datasets/data, but I preferred it kept /xfa:datasets/data

So I modified the code to let it keep the XML namespace prefix. Hope you have some time to review it.

def xpath_soup(element, keep_prefix: bool=True):
    components = []
    child = element if element.name else element.parent
    for parent in child.parents:  # type: bs4.element.Tag
        siblings = parent.find_all(child.name, recursive=False)
        child_name = f'{child.prefix}:{child.name}' if child.prefix and keep_prefix else child.name
        components.append(
            child_name if 1 == len(siblings) else '%s[%d]' % (
                child_name,
                next(i for i, s in enumerate(siblings, 1) if s is child)
                )
            )
        child = parent
    components.reverse()
    return '/%s' % '/'.join(components)

@ergoithz
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Author

@funway thanks for the heads up, snippet updated with your suggestion.

(You might take a look at your code tho, your sibling search wasn't prefixed).

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