I have a regexp and i want to add output of regexp to my url
for exmaple
url = 'blabla.com'
r = re.findall(r'<p>(.*?</a>))
r output - /any_string/on/any/server/
but a dont know how to make get-request with regexp output
blabla.com/any_string/on/any/server/
Don't use regex to parse html. Use a real parser.
I suggest using the lxml.html parser. lxml supports xpath, which is a very powerful way of querying structured documents. There's a ready-to-use make_links_absolute() method that does what you ask. It's also very fast.
As an example, in this question's page HTML source code (the one you're reading now) there's this part:
<li><a id="nav-tags" href="/tags">Tags</a></li>
The xpath expression //a[#id='nav-tags']/#href means: "Get me the href attribute of all <a> tags with id attribute equal to nav-tags". Let's use it:
from lxml import html
url = 'http://stackoverflow.com/questions/3423822/python-url-regexp'
doc = html.parse(url).getroot()
doc.make_links_absolute()
links = doc.xpath("//a[#id='nav-tags']/#href")
print links
The result:
['http://stackoverflow.com/tags']
Just get beautiful soup:
http://www.crummy.com/software/BeautifulSoup/documentation.html#Parsing+a+Document
import urllib2
from BeautifulSoup import BeautifulSoup
page = urllib2.urlopen(url)
soup = BeautifulSoup(page)
soup.findAll('p')
Related
Do you know how to search for specific text inside pythons BeautifulSoup, to find tags - better full path to tags, containing some string?
The common way of BS4 usage is for example:
import requests
from bs4 import BeautifulSoup
url = "https://elementy.ru/novosti_nauki"
website = requests.get(url)
results = BeautifulSoup(website.content, 'html.parser')
and then you can query for all tags with some properties, like header, class, etc.
However I want to go different way, and find the location of the specific text inside this structure?
If you do it with plain HTML text it is really unconvenient.
You could use a css selector or more exactly the Soup Sieve pseudo class :-soup-contains() to search for tags that contains a certain string:
soup.select(':-soup-contains("that")')
or as alternative re.compile():
import re
soup('p', text=re.compile('that')))
Note: If you use soup.find_all(string="that") this wont work, cause it expects to match the exact string.
Example
from bs4 import BeautifulSoup
html = '''
<p>some content</p>
<p>pattern that we like</p>
<p>some content</p>
'''
soup = BeautifulSoup(html, 'html.parser')
soup.select(':-soup-contains("that")')
Output
[<p>pattern that we like</p>]
In python, I'm using the requests module and BS4 to search the web with duckduckgo.com. I went to http://duckduckgo.com/html/?q='hello' manually and got the first results title as <a class="result__a" href="http://example.com"> using the Developer Tools. Now I used the following code to get the href with Python:
html = requests.get('http://duckduckgo.com/html/?q=hello').content
soup = BeautifulSoup4(html, 'html.parser')
result = soup.find('a', class_='result__a')['href']
However, the href looks like gibberish and is completely different from the one i saw manually. ny idea why this is happening?
There are multiple DOM elements with the classname 'result__a'. So, don't expect the first link you see be the first you get.
The 'gibberish' you mentioned is an encoded URL. You'll need to decode and parse it to get the parameters(params) of the URL.
For example:
"/l/?kh=-1&uddg=https%3A%2F%2Fwww.example.com"
The above href contains two params, namely kh and uddg.
uddg is the actual link you need I suppose.
Below code will get all the URL of that particular class, unquoted.
import requests
from bs4 import BeautifulSoup
from urllib.parse import urlparse, parse_qs, unquote
html = requests.get('http://duckduckgo.com/html/?q=hello').content
soup = BeautifulSoup(html, 'html.parser')
for anchor in soup.find_all('a', attrs={'class':'result__a'}):
link = anchor.get('href')
url_obj = urlparse(link)
parsed_url = parse_qs(url_obj.query).get('uddg', '')
if parsed_url:
print(unquote(parsed_url[0]))
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()]
I am new to Python and I am learning it for scraping purposes I am using BeautifulSoup to collect links (i.e href of 'a' tag). I am trying to collect the links under the "UPCOMING EVENTS" tab of site http://allevents.in/lahore/. I am using Firebug to inspect the element and to get the CSS path but this code returns me nothing. I am looking for the fix and also some suggestions for how I can choose proper CSS selectors to retrieve desired links from any site. I wrote this piece of code:
from bs4 import BeautifulSoup
import requests
url = "http://allevents.in/lahore/"
r = requests.get(url)
data = r.text
soup = BeautifulSoup(data)
for link in soup.select( 'html body div.non-overlay.gray-trans-back div.container div.row div.span8 div#eh-1748056798.events-horizontal div.eh-container.row ul.eh-slider li.h-item div.h-meta div.title a[href]'):
print link.get('href')
The page is not the most friendly in the use of classes and markup, but even so your CSS selector is too specific to be useful here.
If you want Upcoming Events, you want just the first <div class="events-horizontal">, then just grab the <div class="title"><a href="..."></div> tags, so the links on titles:
upcoming_events_div = soup.select_one('div.events-horizontal')
for link in upcoming_events_div.select('div.title a[href]'):
print(link['href'])
Note that you should not use r.text; use r.content and leave decoding to Unicode to BeautifulSoup. See Encoding issue of a character in utf-8
import bs4 , requests
res = requests.get("http://allevents.in/lahore/")
soup = bs4.BeautifulSoup(res.text)
for link in soup.select('a[property="schema:url"]'):
print link.get('href')
This code will work fine!!
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')