from bs4 import BeautifulSoup
from fake_useragent import UserAgent
import requests
user = UserAgent()
headers = {
'user-agent' : user.random
}
url = 'https://www.wildberries.ru/?utm_source=domain&utm_campaign=wilberes.ru'
def main():
resp = requests.get(url, headers=headers)
soup = BeautifulSoup(resp.text, 'lxml')
main = soup.find('div', class_='menu-burger__main')
ul = main.find('ul', class_='menu-burger__main-list')
all = ul.find_all_next('li', class_='menu-burger__main-list-item')
f = open('link.txt', 'a')
for lin in all:
get_link = lin.find('a').get('href')
f.write(get_link + '\n')
f.close()
if __name__ == '__main__':
main()
I'm trying to parse a link to a section and its name. I managed to get the link, but how can I get the name if it is not in the tag?
Using the string property seems to be the correct method based on BS4 documentation:
myLink = soup.find('a')
myLinkText = str(myLink.string)
The idea of using str() is to convert the text to regular python strings, as .string returns a BS4 NavigableString object (you may find you don't really have to do this, though - it might be safer though, as well as stripping whitespace from the string so that you don't get weird newlines or padding with the result -- str(myLink.string).strip())
Reference: https://crummy.com/software/BeautifulSoup/bs4/doc/#navigablestring
In your code, you are actually getting the href, so take note that in my code above I am getting the anchor tag, not just it's href attribute.
In this code I think I made a mistake or something because I'm not getting the correct json when I print it, indeed I get nothing but when I index the script I get the json but using .text nothing appears I want the json alone.
CODE :
from bs4 import BeautifulSoup
from urllib.parse import quote_plus
import requests
import selenium.webdriver as webdriver
base_url = 'https://www.instagram.com/{}'
search = input('Enter the instagram account: ')
final_url = base_url.format(quote_plus(search))
response = requests.get(final_url)
print(response.status_code)
if response.ok:
html = response.text
bs_html = BeautifulSoup(html)
scripts = bs_html.select('script[type="application/ld+json"]')
print(scripts[0].text)
Change the line print(scripts[0].text) to print(scripts[0].string).
scripts[0] is a Beautiful Soup Tag object, and its string contents can be accessed through the .string property.
Source: https://www.crummy.com/software/BeautifulSoup/bs4/doc/#string
If you want to then turn the string into a json so that you can access the data, you can do something like this:
...
if response.ok:
html = response.text
bs_html = BeautifulSoup(html)
scripts = bs_html.select('script[type="application/ld+json"]')
json_output = json.loads(scripts[0].string)
Then, for example, if you run print(json_output['name']) you should be able to access the name on the account.
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How can I retrieve the links of a webpage and copy the url address of the links using Python?
Here's a short snippet using the SoupStrainer class in BeautifulSoup:
import httplib2
from bs4 import BeautifulSoup, SoupStrainer
http = httplib2.Http()
status, response = http.request('http://www.nytimes.com')
for link in BeautifulSoup(response, parse_only=SoupStrainer('a')):
if link.has_attr('href'):
print(link['href'])
The BeautifulSoup documentation is actually quite good, and covers a number of typical scenarios:
https://www.crummy.com/software/BeautifulSoup/bs4/doc/
Edit: Note that I used the SoupStrainer class because it's a bit more efficient (memory and speed wise), if you know what you're parsing in advance.
For completeness sake, the BeautifulSoup 4 version, making use of the encoding supplied by the server as well:
from bs4 import BeautifulSoup
import urllib.request
parser = 'html.parser' # or 'lxml' (preferred) or 'html5lib', if installed
resp = urllib.request.urlopen("http://www.gpsbasecamp.com/national-parks")
soup = BeautifulSoup(resp, parser, from_encoding=resp.info().get_param('charset'))
for link in soup.find_all('a', href=True):
print(link['href'])
or the Python 2 version:
from bs4 import BeautifulSoup
import urllib2
parser = 'html.parser' # or 'lxml' (preferred) or 'html5lib', if installed
resp = urllib2.urlopen("http://www.gpsbasecamp.com/national-parks")
soup = BeautifulSoup(resp, parser, from_encoding=resp.info().getparam('charset'))
for link in soup.find_all('a', href=True):
print link['href']
and a version using the requests library, which as written will work in both Python 2 and 3:
from bs4 import BeautifulSoup
from bs4.dammit import EncodingDetector
import requests
parser = 'html.parser' # or 'lxml' (preferred) or 'html5lib', if installed
resp = requests.get("http://www.gpsbasecamp.com/national-parks")
http_encoding = resp.encoding if 'charset' in resp.headers.get('content-type', '').lower() else None
html_encoding = EncodingDetector.find_declared_encoding(resp.content, is_html=True)
encoding = html_encoding or http_encoding
soup = BeautifulSoup(resp.content, parser, from_encoding=encoding)
for link in soup.find_all('a', href=True):
print(link['href'])
The soup.find_all('a', href=True) call finds all <a> elements that have an href attribute; elements without the attribute are skipped.
BeautifulSoup 3 stopped development in March 2012; new projects really should use BeautifulSoup 4, always.
Note that you should leave decoding the HTML from bytes to BeautifulSoup. You can inform BeautifulSoup of the characterset found in the HTTP response headers to assist in decoding, but this can be wrong and conflicting with a <meta> header info found in the HTML itself, which is why the above uses the BeautifulSoup internal class method EncodingDetector.find_declared_encoding() to make sure that such embedded encoding hints win over a misconfigured server.
With requests, the response.encoding attribute defaults to Latin-1 if the response has a text/* mimetype, even if no characterset was returned. This is consistent with the HTTP RFCs but painful when used with HTML parsing, so you should ignore that attribute when no charset is set in the Content-Type header.
Others have recommended BeautifulSoup, but it's much better to use lxml. Despite its name, it is also for parsing and scraping HTML. It's much, much faster than BeautifulSoup, and it even handles "broken" HTML better than BeautifulSoup (their claim to fame). It has a compatibility API for BeautifulSoup too if you don't want to learn the lxml API.
Ian Blicking agrees.
There's no reason to use BeautifulSoup anymore, unless you're on Google App Engine or something where anything not purely Python isn't allowed.
lxml.html also supports CSS3 selectors so this sort of thing is trivial.
An example with lxml and xpath would look like this:
import urllib
import lxml.html
connection = urllib.urlopen('http://www.nytimes.com')
dom = lxml.html.fromstring(connection.read())
for link in dom.xpath('//a/#href'): # select the url in href for all a tags(links)
print link
import urllib2
import BeautifulSoup
request = urllib2.Request("http://www.gpsbasecamp.com/national-parks")
response = urllib2.urlopen(request)
soup = BeautifulSoup.BeautifulSoup(response)
for a in soup.findAll('a'):
if 'national-park' in a['href']:
print 'found a url with national-park in the link'
The following code is to retrieve all the links available in a webpage using urllib2 and BeautifulSoup4:
import urllib2
from bs4 import BeautifulSoup
url = urllib2.urlopen("http://www.espncricinfo.com/").read()
soup = BeautifulSoup(url)
for line in soup.find_all('a'):
print(line.get('href'))
Links can be within a variety of attributes so you could pass a list of those attributes to select.
For example, with src and href attributes (here I am using the starts with ^ operator to specify that either of these attributes values starts with http):
from bs4 import BeautifulSoup as bs
import requests
r = requests.get('https://stackoverflow.com/')
soup = bs(r.content, 'lxml')
links = [item['href'] if item.get('href') is not None else item['src'] for item in soup.select('[href^="http"], [src^="http"]') ]
print(links)
Attribute = value selectors
[attr^=value]
Represents elements with an attribute name of attr whose value is prefixed (preceded) by value.
There are also the commonly used $ (ends with) and * (contains) operators. For a full syntax list see the link above.
Under the hood BeautifulSoup now uses lxml. Requests, lxml & list comprehensions makes a killer combo.
import requests
import lxml.html
dom = lxml.html.fromstring(requests.get('http://www.nytimes.com').content)
[x for x in dom.xpath('//a/#href') if '//' in x and 'nytimes.com' not in x]
In the list comp, the "if '//' and 'url.com' not in x" is a simple method to scrub the url list of the sites 'internal' navigation urls, etc.
just for getting the links, without B.soup and regex:
import urllib2
url="http://www.somewhere.com"
page=urllib2.urlopen(url)
data=page.read().split("</a>")
tag="<a href=\""
endtag="\">"
for item in data:
if "<a href" in item:
try:
ind = item.index(tag)
item=item[ind+len(tag):]
end=item.index(endtag)
except: pass
else:
print item[:end]
for more complex operations, of course BSoup is still preferred.
This script does what your looking for, But also resolves the relative links to absolute links.
import urllib
import lxml.html
import urlparse
def get_dom(url):
connection = urllib.urlopen(url)
return lxml.html.fromstring(connection.read())
def get_links(url):
return resolve_links((link for link in get_dom(url).xpath('//a/#href')))
def guess_root(links):
for link in links:
if link.startswith('http'):
parsed_link = urlparse.urlparse(link)
scheme = parsed_link.scheme + '://'
netloc = parsed_link.netloc
return scheme + netloc
def resolve_links(links):
root = guess_root(links)
for link in links:
if not link.startswith('http'):
link = urlparse.urljoin(root, link)
yield link
for link in get_links('http://www.google.com'):
print link
To find all the links, we will in this example use the urllib2 module together
with the re.module
*One of the most powerful function in the re module is "re.findall()".
While re.search() is used to find the first match for a pattern, re.findall() finds all
the matches and returns them as a list of strings, with each string representing one match*
import urllib2
import re
#connect to a URL
website = urllib2.urlopen(url)
#read html code
html = website.read()
#use re.findall to get all the links
links = re.findall('"((http|ftp)s?://.*?)"', html)
print links
Why not use regular expressions:
import urllib2
import re
url = "http://www.somewhere.com"
page = urllib2.urlopen(url)
page = page.read()
links = re.findall(r"<a.*?\s*href=\"(.*?)\".*?>(.*?)</a>", page)
for link in links:
print('href: %s, HTML text: %s' % (link[0], link[1]))
Here's an example using #ars accepted answer and the BeautifulSoup4, requests, and wget modules to handle the downloads.
import requests
import wget
import os
from bs4 import BeautifulSoup, SoupStrainer
url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/eeg-mld/eeg_full/'
file_type = '.tar.gz'
response = requests.get(url)
for link in BeautifulSoup(response.content, 'html.parser', parse_only=SoupStrainer('a')):
if link.has_attr('href'):
if file_type in link['href']:
full_path = url + link['href']
wget.download(full_path)
I found the answer by #Blairg23 working , after the following correction (covering the scenario where it failed to work correctly):
for link in BeautifulSoup(response.content, 'html.parser', parse_only=SoupStrainer('a')):
if link.has_attr('href'):
if file_type in link['href']:
full_path =urlparse.urljoin(url , link['href']) #module urlparse need to be imported
wget.download(full_path)
For Python 3:
urllib.parse.urljoin has to be used in order to obtain the full URL instead.
BeatifulSoup's own parser can be slow. It might be more feasible to use lxml which is capable of parsing directly from a URL (with some limitations mentioned below).
import lxml.html
doc = lxml.html.parse(url)
links = doc.xpath('//a[#href]')
for link in links:
print link.attrib['href']
The code above will return the links as is, and in most cases they would be relative links or absolute from the site root. Since my use case was to only extract a certain type of links, below is a version that converts the links to full URLs and which optionally accepts a glob pattern like *.mp3. It won't handle single and double dots in the relative paths though, but so far I didn't have the need for it. If you need to parse URL fragments containing ../ or ./ then urlparse.urljoin might come in handy.
NOTE: Direct lxml url parsing doesn't handle loading from https and doesn't do redirects, so for this reason the version below is using urllib2 + lxml.
#!/usr/bin/env python
import sys
import urllib2
import urlparse
import lxml.html
import fnmatch
try:
import urltools as urltools
except ImportError:
sys.stderr.write('To normalize URLs run: `pip install urltools --user`')
urltools = None
def get_host(url):
p = urlparse.urlparse(url)
return "{}://{}".format(p.scheme, p.netloc)
if __name__ == '__main__':
url = sys.argv[1]
host = get_host(url)
glob_patt = len(sys.argv) > 2 and sys.argv[2] or '*'
doc = lxml.html.parse(urllib2.urlopen(url))
links = doc.xpath('//a[#href]')
for link in links:
href = link.attrib['href']
if fnmatch.fnmatch(href, glob_patt):
if not href.startswith(('http://', 'https://' 'ftp://')):
if href.startswith('/'):
href = host + href
else:
parent_url = url.rsplit('/', 1)[0]
href = urlparse.urljoin(parent_url, href)
if urltools:
href = urltools.normalize(href)
print href
The usage is as follows:
getlinks.py http://stackoverflow.com/a/37758066/191246
getlinks.py http://stackoverflow.com/a/37758066/191246 "*users*"
getlinks.py http://fakedomain.mu/somepage.html "*.mp3"
There can be many duplicate links together with both external and internal links. To differentiate between the two and just get unique links using sets:
# Python 3.
import urllib
from bs4 import BeautifulSoup
url = "http://www.espncricinfo.com/"
resp = urllib.request.urlopen(url)
# Get server encoding per recommendation of Martijn Pieters.
soup = BeautifulSoup(resp, from_encoding=resp.info().get_param('charset'))
external_links = set()
internal_links = set()
for line in soup.find_all('a'):
link = line.get('href')
if not link:
continue
if link.startswith('http'):
external_links.add(link)
else:
internal_links.add(link)
# Depending on usage, full internal links may be preferred.
full_internal_links = {
urllib.parse.urljoin(url, internal_link)
for internal_link in internal_links
}
# Print all unique external and full internal links.
for link in external_links.union(full_internal_links):
print(link)
import urllib2
from bs4 import BeautifulSoup
a=urllib2.urlopen('http://dir.yahoo.com')
code=a.read()
soup=BeautifulSoup(code)
links=soup.findAll("a")
#To get href part alone
print links[0].attrs['href']
i can get the links but Idk how to filter https only
To parse HTML use html parser, e.g. BeautifulSoup. To extract desired <a> elements, you can use CSS selector 'a[href^="https"]' (Selects every <a> element whose href attribute value begins with "https"):
import requests
from bs4 import BeautifulSoup
url = 'https://sayamkanwar.com/'
soup = BeautifulSoup(requests.get(url).text, 'lxml')
for a in soup.select('a[href^="https"]'):
print(a['href'])
Prints:
https://sayamkanwar.com/work
https://sayamkanwar.com/about
https://www.facebook.com/sayamkanwar
https://github.com/sayamkanwar
https://codepen.io/sayamk/
https://medium.com/#sayamkanwar/
Further reading:
CSS Selectors Reference
EDIT: Using only builtin modules:
import urllib.request
from html.parser import HTMLParser
url = 'https://sayamkanwar.com/'
class MyHTMLParser(HTMLParser):
def handle_starttag(self, tag, attrs):
if tag=='a':
attrs = dict(attrs)
if 'href' in attrs and attrs['href'].startswith('https'):
print(attrs['href'])
with urllib.request.urlopen(url) as response:
src = response.read().decode('utf-8')
parser = MyHTMLParser()
parser.feed(src)
Prints:
https://sayamkanwar.com/work
https://sayamkanwar.com/about
https://www.facebook.com/sayamkanwar
https://github.com/sayamkanwar
https://codepen.io/sayamk/
https://medium.com/#sayamkanwar/
Try this, I just use request library.
import re
import requests
URL = 'https://sayamkanwar.com/'
response = requests.get(URL)
pattern = r'(a href=")((https):((//)|(\\\\))+([\w\d:##%/;$()~_?\+-=\\\.&](#!)?)*)"'
all_url = re.findall(pattern, response.text)
for url in all_url:
print(url[1])
Output:
https://www.facebook.com/sayamkanwar
https://github.com/sayamkanwar
https://codepen.io/sayamk/
https://medium.com/#sayamkanwar/
Visual output of the regex:
findall() returns empty list when specifying class
Specifying tags work fine
import urllib2
from bs4 import BeautifulSoup
url = "https://www.reddit.com/r/Showerthoughts/top/?sort=top&t=week"
hdr = { 'User-Agent' : 'tempro' }
req = urllib2.Request(url, headers=hdr)
htmlpage = urllib2.urlopen(req).read()
BeautifulSoupFormat = BeautifulSoup(htmlpage,'lxml')
name_box = BeautifulSoupFormat.findAll("a",{'class':'title'})
for data in name_box:
print(data.text)
I'm trying to get only the text of the post. The current code prints out nothing. If I remove the {'class':'title'} it prints out the post text as well as username and comments of the post which I don't want.
I'm using python2 with the latest versions of BeautifulSoup and urllib2
To get all the comments you are going to need a method like selenium which will allow you to scroll. Without that, just to get initial results, you can grab from a script tag in the requests response
import requests
from bs4 import BeautifulSoup as bs
import re
import json
headers = {'User-Agent' : 'Mozilla/5.0'}
r = requests.get('https://www.reddit.com/r/Showerthoughts/top/?sort=top&t=week', headers = headers)
soup = bs(r.content, 'lxml')
script = soup.select_one('#data').text
p = re.compile(r'window.___r = (.*); window')
data = json.loads(p.findall(script)[0])
for item in data['posts']['models']:
print(data['posts']['models'][item]['title'])
The selector you try to use is not good, because you do not have a class = "title" for those posts. Please try this below:
name_box = BeautifulSoupFormat.select('a[data-click-id="body"] > h2')
this finds all the <a data-click-id="body"> where you have <h2> tag that contain the post text you need
More about selectors using BeatufulSoup you can read here:
(https://www.crummy.com/software/BeautifulSoup/bs4/doc/#css-selectors)