Python Http Error 403 Forbidden
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Raise Httperror(req.full_url, Code, Msg, Hdrs, Fp) Urllib.error.httperror: Http Error 403: Forbidden
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Yolk Urllib2.httperror: Http Error 403: Must Access Using Https Instead Of Http
Overflow is a community of 6.2 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up urllib2.HTTPError: HTTP Error 403: Forbidden up vote 41 down vote favorite 25 I am trying to automate download of historic stock data using python. The URL I am trying urllib2 user agent to open responds with a CSV file, but I am unable to open using urllib2. I have tried changing user agent as specified in few questions earlier, I even tried to accept response cookies, with no luck. Can you please help. Note: The same method works for yahoo Finance. Code: import urllib2,cookielib site= "http://www.nseindia.com/live_market/dynaContent/live_watch/get_quote/getHistoricalData.jsp?symbol=JPASSOCIAT&fromDate=1-JAN-2012&toDate=1-AUG-2012&datePeriod=unselected&hiddDwnld=true" hdr = {'User-Agent':'Mozilla/5.0'} req = urllib2.Request(site,headers=hdr) page = urllib2.urlopen(req) Error File "C:\Python27\lib\urllib2.py", line 527, in http_error_default raise HTTPError(req.get_full_url(), code, msg, hdrs, fp) urllib2.HTTPError: HTTP Error 403: Forbidden Thanks for your assistance python http urllib share|improve this question edited Nov 9 '12 at 7:14 Sudar 5,407124790 asked Nov 9 '12 at 6:51 kumar 8491810 Are you use windows as platform ? –Denis Nov 9 '12 at 7:08 add a comment| 2 Answers 2 active oldest votes up vote 77 down vote accepted By adding a few more headers I was able to get the data: import urllib2,cookielib site= "http://www.nseindia.com/live_market/dynaContent/live_watch/get_quote/getHistoricalData.jsp?symbol=JPASSOCIAT&fromDate=1-JAN-2012&toDate=1-AUG-2012&datePeriod=unselected&hid
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Python Urllib Headers
about Stack Overflow the company Business Learn more about hiring developers or posting python user-agent ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack python requests user agent Overflow is a community of 6.2 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up HTTP Error 403: Forbidden up vote -1 down vote favorite 1 http://stackoverflow.com/questions/13303449/urllib2-httperror-http-error-403-forbidden 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. Please help. 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 http://stackoverflow.com/questions/34957748/http-error-403-forbidden 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)) python http python-requests urllib share|improve this question asked Jan 22 at 23:38 Z.Chen 204 If the server is blocking, there's probably not an easy way through. Forbidden means that you are not allowed. –Zizouz212 Jan 22 at 23:41 add a comment| 1 Answer 1 active oldest votes up vote 2 down vote accepted 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
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more http://stackoverflow.com/questions/28396036/python-3-4-urllib-request-error-http-403 about Stack Overflow the company Business Learn more about hiring developers or posting ads https://codedump.io/share/7LcZzKNbi1/1/http-error-403-forbidden-urlib2-python-27 with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 6.2 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up Python 3.4 urllib.request error (http 403) up vote 8 down vote favorite http error I'm trying to open and parse a html page. In python 2.7.8 I have no problem: import urllib url = "https://ipdb.at/ip/66.196.116.112" html = urllib.urlopen(url).read() and everything is fine. However I want to move to python 3.4 and there I get HTTP error 403 (Forbidden). My code: import urllib.request html = urllib.request.urlopen(url) # same URL as before File "C:\Python34\lib\urllib\request.py", line 153, in urlopen return opener.open(url, data, timeout) File "C:\Python34\lib\urllib\request.py", line http error 403 461, in open response = meth(req, response) File "C:\Python34\lib\urllib\request.py", line 574, in http_response 'http', request, response, code, msg, hdrs) File "C:\Python34\lib\urllib\request.py", line 499, in error return self._call_chain(*args) File "C:\Python34\lib\urllib\request.py", line 433, in _call_chain result = func(*args) File "C:\Python34\lib\urllib\request.py", line 582, in http_error_default raise HTTPError(req.full_url, code, msg, hdrs, fp) urllib.error.HTTPError: HTTP Error 403: Forbidden It work for other URLs which don't use https. url = 'http://www.stopforumspam.com/ipcheck/212.91.188.166' is ok. python python-3.x urllib share|improve this question edited Feb 8 '15 at 16:35 falsetru 174k22218268 asked Feb 8 '15 at 15:57 Belial 188110 add a comment| 2 Answers 2 active oldest votes up vote 16 down vote accepted It seems like the site does not like the user agent of Python 3.x. Specifying User-Agent will solve your problem: import urllib.request req = urllib.request.Request(url, headers={'User-Agent': 'Mozilla/5.0'}) html = urllib.request.urlopen(req).read() NOTE Python 2.x urllib version also receives 403 status, but unlike Python 2.x urllib2 and Python 3.x urllib, it does not raise the exception. You can confirm that by following code: print(urllib.urlopen(url).getcode()) # => 403 share|improve this answer edited Feb 8 '15 at 16:43 answered Feb 8 '15 at 16:34 falsetru 174k22218268 Thanks. It worked! –Belial Feb 8 '15 at 16:49 thanks! worked for me too –De
Go Groovy Haml Handlebars Haskell HTML HTTP Ini iOS Jade Java Javascript jQuery JSON Julia Keyman LaTeX Linux Less LOLCODE Makefile Markdown MATLAB MySQL NASM Node.js NSIS Objective-C Pascal Perl PHP PHP Extras PowerShell Python R React JSX reST (reStructuredText) Rip Ruby Rust SAS Sass (Sass) Sass (Scss) Scala Scheme Smalltalk Smarty SQL Stylus Swift Twig TypeScript Vb.net VHDL Wiki markup YAML Other Search FancyDolphin - 7 months ago 229x HTML Question HTTP Error 403: Forbidden urlib2 Python 2.7 I've been successfully been able to use urllib2 but for this website I was testing all of a sudden it didn't work. I've looked on the forum and tried some of the fixes and it doesn't seem to work. Below is an example of one way it was solved but isn't working for me. Can someone help me be able to connect to it. The code that gives the error: from bs4 import BeautifulSoup
import urllib2
proxy_support = urllib2.ProxyHandler({"http":"http://username:password@ip:port"})
hdr = {'Accept': 'text/html,application/xhtml+xml,*/*'}
url = 'http://www.carnextdoor.com.au/'
opener = urllib2.build_opener(proxy_support)
urllib2.install_opener(opener)
req=urllib2.Request(url,headers=hdr)
#Here I get the error with and without using the header or going html = urllib2.urlopen(url).read()
html = urllib2.urlopen(req).read()
soup=BeautifulSoup(html,"html5lib")
print soup
Padraic Cunningham Answer Email {} Share I got a 403 until I added a user-agent, the following was enough to work for me: hdr = {'Accept': 'text/html,application/xhtml+xml,*/*',"user-agent":"Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/48.0.2564.116 Safari/537.36"} url = 'http://www.carnextdoor.com.au/' req=urllib2.Request(url,headers=hdr) #Here I get the error with and without using the header or going html = urllib2.urlopen(url).read() html = urllib2.urlopen(req).read() soup=BeautifulSoup(html,"html5lib") print soup Without user-agent: In [10]: hdr = {'Accept': 'text/html,application/xhtml+xml,*/*'} In [11]: url = 'http://www.carnextdoor.com.au/' In [12]: req=