HTTP Error 403: Forbidden with urlretrieve
Asked Answered
P

1

17

I am trying to download a PDF, however I get the following error: HTTP Error 403: Forbidden

I am aware that the server is blocking for whatever reason, but I cant seem to find a solution.

import urllib.request
import urllib.parse
import requests


def download_pdf(url):

full_name = "Test.pdf"
urllib.request.urlretrieve(url, full_name)


try: 
url =         ('http://papers.xtremepapers.com/CIE/Cambridge%20IGCSE/Mathematics%20(0580)/0580_s03_qp_1.pdf')

print('initialized')

hdr = {}
hdr = {
'User-Agent': 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_11_2)     AppleWebKit/537.36 (KHTML, like Gecko) Chrome/47.0.2526.106 Safari/537.36',
'Content-Length': '136963',
}



print('HDR recieved')

req = urllib.request.Request(url, headers=hdr)

print('Header sent')

resp = urllib.request.urlopen(req)

print('Request sent')

respData = resp.read()

download_pdf(url)


print('Complete')

except Exception as e:
print(str(e))
Paff answered 22/1, 2016 at 23:38 Comment(2)
If the server is blocking, there's probably not an easy way through. Forbidden means that you are not allowed.Dragoman
Possible duplicate of HTTP Error 403: Forbidden while downloading file using urllibLydell
P
43

You seem to have already realised this; the remote server is apparently checking the user agent header and rejecting requests from Python's urllib. But urllib.request.urlretrieve() doesn't allow you to change the HTTP headers, however, you can use urllib.request.URLopener.retrieve():

import urllib.request

opener = urllib.request.URLopener()
opener.addheader('User-Agent', 'whatever')
filename, headers = opener.retrieve(url, 'Test.pdf')

N.B. You are using Python 3 and these functions are now considered part of the "Legacy interface", and URLopener has been deprecated. For that reason you should not use them in new code.

The above aside, you are going to a lot of trouble to simply access a URL. Your code imports requests, but you don't use it - you should though because it is much easier than urllib. This works for me:

import requests

url = 'http://papers.xtremepapers.com/CIE/Cambridge%20IGCSE/Mathematics%20(0580)/0580_s03_qp_1.pdf'
r = requests.get(url)
with open('0580_s03_qp_1.pdf', 'wb') as outfile:
    outfile.write(r.content)
Panel answered 22/1, 2016 at 23:50 Comment(2)
While a good point - It doesn't explain the reason for the 403 error.Dragoman
The question was not asking for reason.Decrement

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