How to extract all links from a website using python

Question

I have written a script to extract links from websites which works fine This is the source Code


import requests
from bs4 import BeautifulSoup
Web=requests.get("https://www.google.com/")
soup=BeautifulSoup(Web.text,'lxml')
for link in soup.findAll('a'):
    print(link['href'])

##Out put
https://www.google.com.sa/imghp?hl=ar&tab=wi
https://maps.google.com.sa/maps?hl=ar&tab=wl
https://www.youtube.com/?gl=SA&tab=w1
https://news.google.com/?tab=wn
https://mail.google.com/mail/?tab=wm
https://drive.google.com/?tab=wo
https://calendar.google.com/calendar?tab=wc
https://www.google.com.sa/intl/ar/about/products?tab=wh
http://www.google.com.sa/history/optout?hl=ar
/preferences?hl=ar
https://accounts.google.com/ServiceLogin?hl=ar&passive=true&continue=https://www.google.com/&ec=GAZAAQ
/search?safe=strict&ie=UTF-8&q=%D9%86%D9%88%D8%B1+%D8%A7%D9%84%D8%B4%D8%B1%D9%8A%D9%81&oi=ddle&ct=174786979&hl=ar&kgmid=/m/0562zv&sa=X&ved=0ahUKEwiq8feoiqDwAhUK8BQKHc7UD7oQPQgD
/advanced_search?hl=ar-SA&authuser=0
https://www.google.com/setprefs?sig=0_mwAqJUgnrqSouOmGk0UvVz7GgkY%3D&hl=en&source=homepage&sa=X&ved=0ahUKEwiq8feoiqDwAhUK8BQKHc7UD7oQ2ZgBCAU
/intl/ar/ads/
http://www.google.com/intl/ar/services/
/intl/ar/about.html
https://www.google.com/setprefdomain?prefdom=SA&prev=https://www.google.com.sa/&sig=K_e_0jdE_IjI-G5o1qMYziPpQwHgs%3D
/intl/ar/policies/privacy/
/intl/ar/policies/terms/

But the problem is that when I change the website to https://www.jarir.com/, it will not work

import requests
from bs4 import BeautifulSoup
Web=requests.get("https://www.jarir.com/")
soup=BeautifulSoup(Web.text,'lxml')
for link in soup.findAll('a'):
    print(link['href'])

#out put
#

The out put will be #

Have you searched for the problem? What were the results? – NoDataDumpNoContribution Apr 28 '21 at 04:58 — NoDataDumpNoContribution, Apr 28 '21 at 04:58

score 6 · Answer 1 · answered Apr 28 '21 at 04:55

I suggest adding a random header function to avoid the website detecting python-requests as the browser/agent. The code below returns all of the links as requested.

Notice the randomization of the headers and how this code uses the headers parameter in the requests.get method.

import requests
from bs4 import BeautifulSoup
from random import choice

desktop_agents = [
    'Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/54.0.2840.99 Safari/537.36',
    'Mozilla/5.0 (Windows NT 10.0; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/54.0.2840.99 Safari/537.36',
    'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/54.0.2840.99 Safari/537.36',
    'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_12_1) AppleWebKit/602.2.14 (KHTML, like Gecko) Version/10.0.1 Safari/602.2.14',
    'Mozilla/5.0 (Windows NT 10.0; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/54.0.2840.71 Safari/537.36',
    'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_12_1) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/54.0.2840.98 Safari/537.36',
    'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_11_6) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/54.0.2840.98 Safari/537.36',
    'Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/54.0.2840.71 Safari/537.36',
    'Mozilla/5.0 (Windows NT 6.1; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/54.0.2840.99 Safari/537.36',
    'Mozilla/5.0 (Windows NT 10.0; WOW64; rv:50.0) Gecko/20100101 Firefox/50.0']


def random_headers():
    return {'User-Agent': choice(desktop_agents),
            'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,*/*;q=0.8'}

Web=requests.get("https://www.jarir.com", headers=random_headers())
soup=BeautifulSoup(Web.text,'lxml')
for link in soup.findAll('a'):
    print(link['href'])

score 2 · Accepted Answer · answered Apr 28 '21 at 04:36

The site is blocked for Python Bots:

<h1>Access denied</h1>
<p>This website is using a security service to protect itself from online attacks.</p>

You can try adding an user agent to your code, like below:

import requests
from bs4 import BeautifulSoup

headers = {'User-Agent': 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_15_7) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/90.0.4430.85 Safari/537.36'}
Web=requests.get("https://www.jarir.com/", headers=headers)
soup=BeautifulSoup(Web.text)
for link in soup.findAll('a'):
    print(link['href'])

The output is something like:

https://www.jarir.com/wishlist/
https://www.jarir.com/sales/order/history/
https://www.jarir.com/afs/
https://www.jarir.com/contacts/
tel:+966920000089
/cdn-cgi/l/email-protection#6300021106230902110a114d000c0e
https://www.jarir.com/faq/
https://www.jarir.com/warranty_policy/
https://www.jarir.com/return_exchange/
https://www.jarir.com/contacts/
https://www.jarir.com/terms-of-service/
https://www.jarir.com/privacy-policy/
https://www.jarir.com/storelocator/

How to extract all links from a website using python

2 Answers2