Develop Reference

How to select a tags and scrape href value?

I am having trouble getting hyperlinks for tennis matches listed on a webpage, how do I go about fixing the code below so that it can obtain it through print?
import requests
from bs4 import BeautifulSoup
response = requests.get("https://www.betexplorer.com/results/tennis/?year=2022&month=11&day=02")
webpage = response.content
soup = BeautifulSoup(webpage, "html.parser")
print(soup.findAll('a href'))

In newer code avoid old syntax findAll() instead use find_all() or select() with css selectors - For more take a minute to check docs
Select your elements more specific and use set comprehension to avoid duplicates:
set('https://www.betexplorer.com'+a.get('href') for a in soup.select('a[href^="/tennis"]:has(strong)'))
Example
import requests
from bs4 import BeautifulSoup
r = requests.get('https://www.betexplorer.com/results/tennis/?year=2022&month=11&day=02')
soup = BeautifulSoup(r.text)
set('https://www.betexplorer.com'+a.get('href') for a in soup.select('a[href^="/tennis"]:has(strong)'))
Output
{'https://www.betexplorer.com/tennis/itf-men-singles/m15-new-delhi-2/sinha-nitin-kumar-vardhan-vishnu/tOasQaJm/',
'https://www.betexplorer.com/tennis/itf-women-doubles/w25-jerusalem/mushika-mao-mushika-mio-cohen-sapir-nagornaia-sofiia/xbNOHTEH/',
'https://www.betexplorer.com/tennis/itf-men-singles/m25-jakarta-2/barki-nathan-anthony-sun-fajing/zy2r8bp0/',
'https://www.betexplorer.com/tennis/itf-women-singles/w15-solarino/margherita-marcon-abbagnato-anastasia/lpq2YX4d/',
'https://www.betexplorer.com/tennis/itf-women-singles/w60-sydney/lee-ya-hsuan-namigata-junri/CEQrNPIG/',
'https://www.betexplorer.com/tennis/itf-men-doubles/m15-sharm-elsheikh-16/echeverria-john-marrero-curbelo-ivan-ianin-nikita-jasper-lai/nsGbyqiT/',...}

Change the last line to
print([a['href'] for a in soup.findAll('a')])
See a full tutorial here: https://pythonprogramminglanguage.com/get-links-from-webpage/

BeautifullSoup returns the whole DIV, but without the value

This might be a bit of a basic question, but either I do not know how to phrase it, or I'm not finding the answer.
So, I want to scrape a specific value of a website (18.73kWh) in this scenario.
> <div class="itemized-bill-header-consumption"data-bind="text:$root.formatItemizedbillConsumption(key.consumption,key.type)">18.73kWh</div>
So I am using Python and BeutifullSoup to get the value,
kwh = soup.findAll('div',{"class":"itemized-bill-header-consumption"})
The thing is, that as a result, i'm getting
[<div class="itemized-bill-header-consumption" data-bind="text:$root.formatItemizedbillConsumption(key.consumption,key.type)"></div>]
Which is pretty much everything minus the value I want... and I can't figure out why.
Thanks in advance for your help

Use the get_text() method.
html = """
<div class="itemized-bill-header-consumption"data-bind="text:$root.formatItemizedbillConsumption(key.consumption,key.type)">18.73kWh</div>
"""
from bs4 import BeautifulSoup
soup = BeautifulSoup(html, features='lxml')
for div in soup.findAll('div',{"class":"itemized-bill-header-consumption"}):
print(div.get_text())
Output
18.73kWh

You can use CSS selector select. You can try it:
from bs4 import BeautifulSoup
html_doc="""<div class="itemized-bill-header-consumption"data-bind="text:$root.formatItemizedbillConsumption(key.consumption,key.type)">18.73kWh</div>"""
soup = BeautifulSoup(html_doc, 'lxml')
kwh = soup.select("div", class_="itemized-bill-header-consumption")[0].text
print(kwh)
Output will be:
18.73kWh

Python Beautiful Soup - find value based on text in HTML

I am having a problem finding a value in a soup based on text. Here is the code
from bs4 import BeautifulSoup as bs
import requests
import re
html='http://finance.yahoo.com/q/ks?s=aapl+Key+Statistics'
r = requests.get(html)
soup = bs(r.text)
findit=soup.find("td", text=re.compile('Market Cap'))
This returns [], yet there absolutely is text in a 'td' tag with 'Market Cap'.
When I use
soup.find_all("td")
I get a result set which includes:
<td class="yfnc_tablehead1" width="74%">Market Cap (intraday)<font size="-1"><sup>5</sup></font>:</td>

Explanation:
The problem is that this particular tag has other child elements and the .string value, which is checked when you apply the text argument, is None (bs4 has it documented here).
Solutions/Workarounds:
Don't specify the tag name here at all, find the text node and go up to the parent:
soup.find(text=re.compile('Market Cap')).parent.get_text()
Or, you can use find_parent() if td is not the direct parent of the text node:
soup.find(text=re.compile('Market Cap')).find_parent("td").get_text()
You can also use a "search function" to search for the td tags and see if the direct text child nodes has the Market Cap text:
soup.find(lambda tag: tag and
tag.name == "td" and
tag.find(text=re.compile('Market Cap'), recursive=False))
Or, if you are looking to find the following number 5:
soup.find(text=re.compile('Market Cap')).next_sibling.get_text()

You can't use regex with tag. It just won't work. Don't know if it's a bug of specification. I just search after all, and then get the parent back in a list comprehension cause "td" "regex" would give you the td tag.
Code
from bs4 import BeautifulSoup as bs
import requests
import re
html='http://finance.yahoo.com/q/ks?s=aapl+Key+Statistics'
r = requests.get(html)
soup = bs(r.text, "lxml")
findit=soup.find_all(text=re.compile('Market Cap'))
findit=[x.parent for x in findit if x.parent.name == "td"]
print(findit)
Output
[<td class="yfnc_tablehead1" width="74%">Market Cap (intraday)<font size="-1"><sup>5</sup></font>:</td>]

Regex is just a terrible thing to integrate into parsing code and in my humble opinion should be avoided whenever possible.
Personally, I don't like BeautifulSoup due to its lack of XPath support. What you're trying to do is the sort of thing that XPath is ideally suited for. If I were doing what you're doing, I would use lxml for parsing rather than BeautifulSoup's built in parsing and/or regex. It's really quite elegant and extremely fast:
from lxml import etree
import requests
source = requests.get('http://finance.yahoo.com/q/ks?s=aapl+Key+Statistics').content
parsed = etree.HTML(source)
tds_w_market_cap = parsed.xpath('//td[contains(., "Market Cap")]')
FYI the above returns an lxml object rather than the text of the page source. In lxml you don't really work with the source directly, per se. If you need to return a list of the actual source for some reason, you would add something like:
print [etree.tostring(i) for i in tds_w_market_cap]
If you absolutely have to use BeautifulSoup for this task, then I'd use a list comprehension:
from bs4 import BeautifulSoup as bs
import requests
source = requests.get('http://finance.yahoo.com/q/ks?s=aapl+Key+Statistics').content
parsed = bs(source, 'lxml')
tds_w_market_cap = [i for i in parsed.find_all('td') if 'Market Cap' in i.get_text()]

Enquiry: Beautiful soup + a href pattern not scraping like I want

I have the following html pattern I want to scrap using BeautifulSoup. The html pattern is:
TITLE
I want to grab TITLE and the information that is displayed in the link. That is, if you clicked the link there is a a description of the TITLE. I want that description.
I started with just trying to grab the title with the following code:
import urllib
from bs4 import BeautifulSoup
import re
webpage = urrlib.urlopen("http://urlofinterest")
title = re.compile('<a>(.*)</a>')
findTitle = re.findall(title,webpage)
print findTile
My output is:
% python beta2.py
[]
So this is obviously not even finding the title. I even tried <a href>(.*)</a> and that didn't work. Based on my reading of the documentation and I thought BeautifulSoup will grab whatever text is between the symbols I give it. In this case, , so what am I doing wrong?

How come you're importing beautifulsoup and then not using it at all?
webpage = urrlib.urlopen("http://urlofinterest")
You'll want to read the data from this, so that:
webpage = urrlib.urlopen("http://urlofinterest").read()
Something like (should get you to a point to go further):
>>> blah = 'TITLE'
>>> from bs4 import BeautifulSoup
>>> soup = BeautifulSoup(blah) # change to webpage later
>>> for tag in soup('a', href=True):
print tag['href'], tag.string
link TITLE

can we use XPath with BeautifulSoup?

I am using BeautifulSoup to scrape an URL and I had the following code, to find the td tag whose class is 'empformbody':
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, please provide me example code.

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.

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

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.

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]

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

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

I've searched through their docs and it seems there is no 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.

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>More information...</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'))

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()

use soup.find(class_='myclass')