Python 3.5 urllib.request 403 Forbidden Error

Question

import urllib.request
import urllib
from bs4 import BeautifulSoup


url = "https://www.brightscope.com/ratings"
page = urllib.request.urlopen(url)
soup = BeautifulSoup(page, "html.parser")

print(soup.title)

I was trying to go to the above site and the code keeps spitting out a 403 Forbidden Error.

Any Ideas?

C:\Users\jerem\AppData\Local\Programs\Python\Python35-32\python.exe "C:/Users/jerem/PycharmProjects/webscraper/url scraper.py" Traceback (most recent call last): File "C:/Users/jerem/PycharmProjects/webscraper/url scraper.py", line 7, in page = urllib.request.urlopen(url) File "C:\Users\jerem\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 163, in urlopen return opener.open(url, data, timeout) File "C:\Users\jerem\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 472, in open response = meth(req, response) File "C:\Users\jerem\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 582, in http_response 'http', request, response, code, msg, hdrs) File "C:\Users\jerem\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 510, in error return self._call_chain(*args) File "C:\Users\jerem\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 444, in _call_chain result = func(*args) File "C:\Users\jerem\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 590, in http_error_default raise HTTPError(req.full_url, code, msg, hdrs, fp) urllib.error.HTTPError: HTTP Error 403: Forbidden

Answer 1

import requests
from bs4 import BeautifulSoup


url = "https://www.brightscope.com/ratings"
headers = {'User-Agent':'Mozilla/5.0'}
page = requests.get(url)
soup = BeautifulSoup(page.text, "html.parser")

print(soup.title)

out:

<title>BrightScope Ratings</title>

First, use requests rather than urllib.

Than, add headers to requests, if not, the site will ban your, because the default User-Agent is crawler, which the site do not like.