CODEWITHSUNDEEP
pythonweb-scrapingbeautifulsoup
V.Anh
V.Anh

Reputation: 535

Access denied while scraping

I want to create a script to go on to https://www.size.co.uk/featured/footwear/ and scrape the content but somehow when i run the script, i got access denied. Here is the code:

from urllib import urlopen
from bs4 import BeautifulSoup as BS
url = urlopen('https://www.size.co.uk/')
print BS(url, 'lxml')

The output is 

<html><head>
<title>Access Denied</title>
</head><body>
<h1>Access Denied</h1>

You don't have permission to access "http://www.size.co.uk/" on this server.
<p>
Reference #18.6202655f.1498945327.11002828
</p></body>
</html>

When i try it with other websites, the code works fine and also when i use Selenium, nothing happens but i still want to know how to bypass this error without using Selenium. But when i use Selenium on different website like http://www.footpatrol.co.uk/shop i got the same Access Denied error, here is the code for footpatrol:

from selenium import webdriver

driver = webdriver.PhantomJS('C:\Users\V\Desktop\PY\web_scrape\phantomjs.exe')
driver.get('http://www.footpatrol.com')
pageSource = driver.page_source
soup = BS(pageSource, 'lxml')
print soup

Output is:

<html><head>
<title>Access Denied</title>
</head><body>
<h1>Access Denied</h1>

You don't have permission to access "http://www.footpatrol.co.uk/" on this 
server.<p>
Reference #18.6202655f.1498945644.110590db


</p></body></html>

Upvotes: 2

Views: 23835

Answers (2)

mamal
mamal

Reputation: 1986

try this :

  headers = {'User-Agent': 'Mozilla/5.0 (Windows NT 6.1; WOW64; rv:50.0) Gecko/20100101 
  Firefox/50.0'}
  source=requests.get(url, headers=headers).text
  print(source)

Upvotes: 2

Dmitriy Fialkovskiy
Dmitriy Fialkovskiy

Reputation: 3225

import requests
from bs4 import BeautifulSoup as BS

url = 'https://www.size.co.uk/'
agent = {"User-Agent":'Mozilla/5.0 (Windows NT 6.3; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/59.0.3071.115 Safari/537.36'}
page = requests.get(url, headers=agent)
print (BS(page.content, 'lxml'))

Upvotes: 9

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