I am using urllib.urlretrieve in Python to download websites. Though some websites seem to not want me to download them, unless they have a proper referrer from their own site. Does anybody know of a way I can set a referrer in one of Python's libraries or a external one to.
import urllib2
req = urllib2.Request('http://www.example.com/')
req.add_header('Referer', 'http://www.python.org/')
r = urllib2.urlopen(req)
adopted from http://docs.python.org/library/urllib2.html
urllib makes it hard to send arbitrary headers with the request; you could use urllib2, which lets you build and send a Request object with arbitrary headers (including of course the -- alas sadly spelled;-) -- Referer). Doesn't offer urlretrieve, but it's easy to just urlopen as you with and copy the resulting file-like object to disk if you want (directly, or e.g. via shutil functions).
Also, using urllib2 with build_opener you can do this:
import urllib2
opener = urllib2.build_opener()
opener.addheaders = [('Referer', 'http://www.python.org/')]
opener.open('http://www.example.com/')
Related
I'm attempting to connect to a website that requires you to have a specific cookie to access it. For the sake of this question, we'll call the cookie 'required_cookie' and the value 'required_value'.
This is my code:
import urllib
import http.cookiejar
cj = http.cookiejar.CookieJar()
opener = urllib.request.build_opener(urllib.request.HTTPCookieProcessor(cj))
opener.addheaders = [('required_cookie', 'required_value'), ('User-Agent', 'Mozilla/5.0')]
urllib.request.install_opener(opener)
req = Request('https://www.thewebsite.com/')
webpage = urlopen(req).read()
print(webpage)
I'm new to urllib so please answer me as a beginner
To do this with urllib, you need to:
Construct a Cookie object. The constructor isn't documented in the docs, but if you help(http.cookiejar.Cookie) in the interactive interpreter, you can see that its constructor demands values for all 16 attributes. Notice that the docs say, "It is not expected that users of http.cookiejar construct their own Cookie instances."
Add it to the cookiejar with cj.set_cookie(cookie).
Tell the cookiejar to add the correct headers to the request with cj.add_cookie_headers(req).
Assuming you've configured the policy correctly, you're set.
But this is a huge pain. As the docs for urllib.request say:
See also The Requests package is recommended for a higher-level HTTP client interface.
And, unless you have some good reason you can't install requests, you really should go that way. urllib is tolerable for really simple cases, and it can be handy when you need to get deep under the covers—but for everything else, requests is much better.
With requests, your whole program becomes a one-liner:
webpage = requests.get('https://www.thewebsite.com/', cookies={'required_cookie': required_value}, headers={'User-Agent': 'Mozilla/5.0'}).text
… although it's probably more readable as a few lines:
cookies = {'required_cookie': required_value}
headers = {'User-Agent': 'Mozilla/5.0'}
response = requests.get('https://www.thewebsite.com/', cookies=cookies, headers=headers)
webpage = response.text
With the help of Kite documentation: https://www.kite.com/python/answers/how-to-add-a-cookie-to-an-http-request-using-urllib-in-python
You can add cookie this way:
import urllib
a_request = urllib.request.Request("http://www.kite.com/")
a_request.add_header("Cookie", "cookiename=cookievalue")
or in a different way:
from urllib.request import Request
url = "https://www.kite.com/"
req = Request(url, headers={'User-Agent': 'Mozilla/5.0', 'Cookie':'myCookie=lovely'})
Sometimes it's necessary to change the version attribute when retrieving a request using FancyURLopener, e.g.
from urllib.request import FancyURLopener
class NewOpener(FancyURLopener):
version = 'Some fancy thing'
url = 'www.google.com'
opener = NewOpener.retrieve(url, 'google.html')
Is there an equivalence in the requests library when using requests.get()?
As #Sraw commented, the "version" is basically the user-agent file in the header, so
requests.get(url, headers={'User-agent': 'Some fancy thing'}
Using urlopen also for url queries seems obvious. What I tried is:
import urllib2
query='http://www.onvista.de/aktien/snapshot.html?ID_OSI=86627'
f = urllib2.urlopen(query)
s = f.read()
f.close()
However, for this specific url query it fails with HTTP error 403 forbidden
When entering this query in my browser, it works.
Also when using http://www.httpquery.com/ to submit the query, it works.
Do you have suggestions how to use Python right to grab the correct response?
Looks like it requires cookies... (which you can do with urllib2), but an easier way if you're doing this, is to use requests
import requests
session = requests.session()
r = session.get('http://www.onvista.de/aktien/snapshot.html?ID_OSI=86627')
This is generally a much easier and less-stressful method of retrieving URLs in Python.
requests will automatically store and re-use cookies for you. Creating a session is slightly overkill here, but is useful for when you need to submit data to login pages etc..., or re-use cookies across a site... etc...
using urllib2 is something like
import urllib2, cookielib
cookies = cookielib.CookieJar()
opener = urllib2.build_opener( urllib2.HTTPCookieProcessor(cookies) )
data = opener.open('url').read()
It appears that the urllib2 default user agent is banned by the host. You can simply supply your own user agent string:
import urllib2
url = 'http://www.onvista.de/aktien/snapshot.html?ID_OSI=86627'
request = urllib2.Request(url, headers={"User-Agent" : "MyUserAgent"})
contents = urllib2.urlopen(request).read()
print contents
Using urllib2 and trying to get an https page, it keeps failing with
Invalid url, unable to resolve
The url is
https://www.domainsbyproxy.com/default.aspx
but I have this happening on multiple https sites.
I am using python 2.7, and below is the code I am using to setup the connection
opener = urllib2.OpenerDirector()
opener.add_handler(urllib2.HTTPHandler())
opener.add_handler(urllib2.HTTPDefaultErrorHandler())
opener.addheaders = [('Accept-encoding', 'gzip')]
fetch_timeout = 12
response = opener.open(url, None, fetch_timeout)
The reason I am setting handlers manually is because I don't want redirects handled (which works fine). The above works fine for http requests, however https - fails.
Any clues?
You should be using HTTPSHandler instead of HTTPHandler
If you don't mind external libraries, consider the excellent requests module. It takes care of these quirks with urllib.
Your code, using requests is:
import requests
r = requests.get(url, headers={'Accept-encoding': 'gzip'}, timeout=12)
how do i change my headers and request so that i appear as firefox ...
like when request to some servers
import urllib
f = urllib.urlopen("rss feed")
they deny my request saying your client dosent have permission...
i get reply but the reply contains " your client dosent have permission"
so how do i get around this and get the data...
http://vsbabu.org/mt/archives/2003/05/27/urllib2_setting_http_headers.html
If you want to use good old urllib instead of newer, fancier urllib2, then as urllib's docs say, and I quote,
For example, applications may want to specify a different User-Agent header than URLopener defines. This can be accomplished with the following code:
import urllib
class AppURLopener(urllib.FancyURLopener):
version = "App/1.7"
urllib._urlopener = AppURLopener()
Of course, you'll want a version (aka User-Agent header) suitable for whatever version of Firefox (or w/ever else;-) you want to pretend you are;-).