BeautifulSoup提取XPATH或CSS节点的路径

Question

我想从HTML中提取一些数据,然后能够在客户端突出显示提取的元素,而无需修改源HTML.而XPath或CSS Path看起来很棒.可以直接从BeautifulSoup中提取XPATH或CSS路径吗？
现在我使用target元素的标记,然后使用lxml lib来提取xpath,这对性能非常不利.我知道BSXPath.py- 它不适用于BS4.由于复杂性,重写所有使用本机lxml lib的解决方案是不可接受的.

import bs4
import cStringIO
import random
from lxml import etree


def get_xpath(soup, element):
  _id = random.getrandbits(32)
  for e in soup():
    if e == element:
      e['data-xpath'] = _id
      break
  else:
    raise LookupError('Cannot find {} in {}'.format(element, soup))
  content = unicode(soup)
  doc = etree.parse(cStringIO.StringIO(content), etree.HTMLParser())
  element = doc.xpath('//*[@data-xpath="{}"]'.format(_id))
  assert len(element) == 1
  element = element[0]
  xpath = doc.getpath(element)
  return xpath

soup = bs4.BeautifulSoup('<div id=i>hello, <b id=i test=t>world!</b></div>')
xpath = get_xpath(soup, soup.div.b)
assert '//html/bodydiv/b' == xpath

Answer 1

实际上很容易提取简单的CSS/XPath.这与lxml lib相同.

def get_element(node):
  # for XPATH we have to count only for nodes with same type!
  length = len(list(node.previous_siblings)) + 1
  if (length) > 1:
    return '%s:nth-child(%s)' % (node.name, length)
  else:
    return node.name

def get_css_path(node):
  path = [get_element(node)]
  for parent in node.parents:
    if parent.name == 'body':
      break
    path.insert(0, get_element(parent))
  return ' > '.join(path)

soup = bs4.BeautifulSoup('<div></div><div><strong><i>bla</i></strong></div>')
assert get_css_path(soup.i) == 'div:nth-child(2) > strong > i'