Question

[Solved] can we use xpath with BeautifulSoup?

I am using BeautifulSoup to scrape a url and I had the following code

import urllib
import urllib2
from BeautifulSoup import BeautifulSoup

url =  "http://www.example.com/servlet/av/ResultTemplate=AVResult.html"
req = urllib2.Request(url)
response = urllib2.urlopen(req)
the_page = response.read()
soup = BeautifulSoup(the_page)
soup.findAll('td',attrs={'class':'empformbody'})

Now in the above code we can use findAll to get tags and information related to them, but I want to use xpath. Is it possible to use xpath with BeautifulSoup? If possible, can anyone please provide me an example code so that it will be more helpful?

Solution #1:

Nope, BeautifulSoup, by itself, does not support XPath expressions.

An alternative library, lxml, does support XPath 1.0. It has a BeautifulSoup compatible mode where it’ll try and parse broken HTML the way Soup does. However, the default lxml HTML parser does just as good a job of parsing broken HTML, and I believe is faster.

Once you’ve parsed your document into an lxml tree, you can use the .xpath() method to search for elements.

try:
    # Python 2
    from urllib2 import urlopen
except ImportError:
    from urllib.request import urlopen
from lxml import etree

url =  "http://www.example.com/servlet/av/ResultTemplate=AVResult.html"
response = urlopen(url)
htmlparser = etree.HTMLParser()
tree = etree.parse(response, htmlparser)
tree.xpath(xpathselector)

There is also a dedicated lxml.html() module with additional functionality.

Note that in the above example I passed the response object directly to lxml, as having the parser read directly from the stream is more efficient than reading the response into a large string first. To do the same with the requests library, you want to set stream=True and pass in the response.raw object after enabling transparent transport decompression:

import lxml.html
import requests

url =  "http://www.example.com/servlet/av/ResultTemplate=AVResult.html"
response = requests.get(url, stream=True)
response.raw.decode_content = True
tree = lxml.html.parse(response.raw)

Of possible interest to you is the CSS Selector support; the CSSSelector class translates CSS statements into XPath expressions, making your search for td.empformbody that much easier:

from lxml.cssselect import CSSSelector

td_empformbody = CSSSelector('td.empformbody')
for elem in td_empformbody(tree):
    # Do something with these table cells.

Coming full circle: BeautifulSoup itself does have very complete CSS selector support:

for cell in soup.select('table#foobar td.empformbody'):
    # Do something with these table cells.
Respondent: Martijn Pieters

Solution #2:

I can confirm that there is no XPath support within Beautiful Soup.

Solution #3:

As others have said, BeautifulSoup doesn’t have xpath support. There are probably a number of ways to get something from an xpath, including using Selenium. However, here’s a solution that works in either Python 2 or 3:

from lxml import html
import requests

page = requests.get('http://econpy.pythonanywhere.com/ex/001.html')
tree = html.fromstring(page.content)
#This will create a list of buyers:
buyers = tree.xpath('//div[@title="buyer-name"]/text()')
#This will create a list of prices
prices = tree.xpath('//span[@class="item-price"]/text()')

print('Buyers: ', buyers)
print('Prices: ', prices)

I used this as a reference.

Respondent: wordsforthewise

Solution #4:

BeautifulSoup has a function named findNext from current element directed childern,so:

father.findNext('div',{'class':'class_value'}).findNext('div',{'id':'id_value'}).findAll('a') 

Above code can imitate the following xpath:

div[class=class_value]/div[id=id_value]
Respondent: user3820561

Solution #5:

from lxml import etree
from bs4 import BeautifulSoup
soup = BeautifulSoup(open('path of your localfile.html'),'html.parser')
dom = etree.HTML(str(soup))
print dom.xpath('//*[@id="BGINP01_S1"]/section/div/font/text()')

Above used the combination of Soup object with lxml and one can extract the value using xpath

Respondent: Deepak rkm

Solution #6:

when you use lxml all simple:

tree = lxml.html.fromstring(html)
i_need_element = tree.xpath('//a[@class="shared-components"]/@href')

but when use BeautifulSoup BS4 all simple too:

  • first remove “//” and “@”
  • second – add star before “=”

try this magic:

soup = BeautifulSoup(html, "lxml")
i_need_element = soup.select ('a[class*="shared-components"]')

as you see, this does not support sub-tag, so i remove “/@href” part

Solution #7:

I’ve searched through their docs and it seems there is not xpath option. Also, as you can see here on a similar question on SO, the OP is asking for a translation from xpath to BeautifulSoup, so my conclusion would be – no, there is no xpath parsing available.

Respondent: Nikola

Solution #8:

Maybe you can try the following without XPath

from simplified_scrapy.simplified_doc import SimplifiedDoc 
html = '''
<html>
<body>
<div>
    <h1>Example Domain</h1>
    <p>This domain is for use in illustrative examples in documents. You may use this
    domain in literature without prior coordination or asking for permission.</p>
    <p><a href="https://www.iana.org/domains/example">More information...</a></p>
</div>
</body>
</html>
'''
# What XPath can do, so can it
doc = SimplifiedDoc(html)
# The result is the same as doc.getElementByTag('body').getElementByTag('div').getElementByTag('h1').text
print (doc.body.div.h1.text)
print (doc.div.h1.text)
print (doc.h1.text) # Shorter paths will be faster
print (doc.div.getChildren())
print (doc.div.getChildren('p'))
Respondent: dabingsou

Solution #9:

This is a pretty old thread, but there is a work-around solution now, which may not have been in BeautifulSoup at the time.

Here is an example of what I did. I use the “requests” module to read an RSS feed and get its text content in a variable called “rss_text”. With that, I run it thru BeautifulSoup, search for the xpath /rss/channel/title, and retrieve its contents. It’s not exactly XPath in all its glory (wildcards, multiple paths, etc.), but if you just have a basic path you want to locate, this works.

from bs4 import BeautifulSoup
rss_obj = BeautifulSoup(rss_text, 'xml')
cls.title = rss_obj.rss.channel.title.get_text()
Respondent: David A

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