Hy!
I tried to open web-page, that is normally opening in browser, but python just swears and does not want to work.
import urllib.request, urllib.error
f = urllib.request.urlopen('http://www.booking.com/reviewlist.html?cc1=tr;pagename=sapphire')
And another way
import urllib.request, urllib.error
opener=urllib.request.build_opener()
f=opener.open('http://www.booking.com/reviewlist.html?cc1=tr;pagename=sapphi
re')
Both options give one type of error:
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "C:\Python34\lib\urllib\request.py", line 461, in open
response = meth(req, response)
File "C:\Python34\lib\urllib\request.py", line 571, in http_response
'http', request, response, code, msg, hdrs)
File "C:\Python34\lib\urllib\request.py", line 493, in error
result = 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 676, in http_error_302
return self.parent.open(new, timeout=req.timeout)
File "C:\Python34\lib\urllib\request.py", line 461, in open
response = meth(req, response)
File "C:\Python34\lib\urllib\request.py", line 571, 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 579, in http_error_default
raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 400: Bad Request
Any ideas?
They are probably blocking the fact that it isn't coming from a browser. You probably need a valid User-Agent header or something.
Using requests, this works:
import requests
headers =
{
'User-Agent': 'Mozilla/5.0 (Windows NT 6.3; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/37.0.2049.0 Safari/537.36'
}
r = requests.get('http://www.booking.com/reviewlist.html?cc1=tr;pagename=sapphire', headers=headers)
print r
print r.headers
This URL seems to be doing user agent string checking. If I adjust my user agent string in Firefox to Python-urllib/2.7, it fails with the Bad Request you are seeing.
As you are using urllib, you can adjust the User Agent following this tutorial
from urllib.request import FancyURLopener
class MyOpener(FancyURLopener):
version = 'My new User-Agent' # Set this to a string you want for your user agent
myopener = MyOpener()
page = myopener.open('http://www.booking.com/reviewlist.html?cc1=tr;pagename=sapphire')
Related
I want to download images from an URL link which has a random component in it, so i have generated a code to do the same, but i'm getting an error -
Code:
import urllib.request
import random
random_number=random.randint(500,600)
url_image="'https://csgostash.com/img/skins/s"+str(random_number)+"fn.png'"
image=urllib.request.urlretrieve(url_image, 'skin.png')
Error:
Traceback (most recent call last):
File "C:/Users/luke/Desktop/scraper/test image download/cs test.py", line 8, in <module>
image=urllib.request.urlretrieve(url_image, 'skin.png')
File "C:\Users\luke\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 187, in urlretrieve
with contextlib.closing(urlopen(url, data)) as fp:
File "C:\Users\luke\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 162, in urlopen
return opener.open(url, data, timeout)
File "C:\Users\luke\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 465, in open
response = self._open(req, data)
File "C:\Users\luke\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 488, in _open
'unknown_open', req)
File "C:\Users\luke\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 443, in _call_chain
result = func(*args)
File "C:\Users\luke\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 1310, in unknown_open
raise URLError('unknown url type: %s' % type)
urllib.error.URLError: <urlopen error unknown url type: 'https>
First, url_image has an weird syntax.
url_image="https://csgostash.com/img/skins/s"+str(random_number)+"fn.png"
If you fix this, you will notice an 403 - Vax! Protection against bot: use a user agent.
import urllib.request
import random
random_number=random.randint(500,600)
url_image="https://csgostash.com/img/skins/s"+str(random_number)+"fn.png"
user_agent = 'Mozilla/5.0 (Windows NT 6.1; Win64; x64)'
headers = {'User-Agent': user_agent}
req = urllib.request.Request(url_image, None, headers)
print(url_image)
#image, h = urllib.request.urlretrieve(url_image)
with urllib.request.urlopen(req) as response:
the_page = response.read()
print (the_page)
Edit: ofcourse you may save to file:
with open('skin.png', 'wb') as f:
f.write(the_page)
Check out this project using requests.
I was trying to get a web page, but got into this problem. I've looked up some references, and this is what I've done so far:
import sys
import urllib2
from bs4 import BeautifulSoup
user = 'myuserID'
password = "mypassword"
ip = sys.argv[1]
url = "http://www.websites.com/" + ip
passman = urllib2.HTTPPasswordMgrWithDefaultRealm()
passman.add_password(None, url, user, password)
handler = urllib2.HTTPBasicAuthHandler(passman)
opener = urllib2.build_opener(handler)
urllib2.install_opener(opener)
header = {
'Connection' : 'keep-alive',
'User-Agent' : 'Mozilla/5.0 (Windows NT 6.1; WOW64; rv:38.0) Gecko/20100101 Firefox/38.0',
'Accept-Language' : 'en-US,en;q=0.5',
'Accept-Encoding' : 'gzip, deflate'
}
html = urllib2.urlopen(urllib2.Request(url, None, header))
soup = BeautifulSoup(html, 'html.parser')
# some if else function afterwards #
When I try to run the script, it shows this kind of error:
python checker.py 8.8.8.8
Traceback (most recent call last):
File "checker.py", line 34, in <module>
html = urllib2.urlopen(urllib2.Request(url, None, header))
File "C:\Python27\lib\urllib2.py", line 154, in urlopen
return opener.open(url, data, timeout)
File "C:\Python27\lib\urllib2.py", line 437, in open
response = meth(req, response)
File "C:\Python27\lib\urllib2.py", line 550, in http_response
'http', request, response, code, msg, hdrs)
File "C:\Python27\lib\urllib2.py", line 469, in error
result = self._call_chain(*args)
File "C:\Python27\lib\urllib2.py", line 409, in _call_chain
result = func(*args)
File "C:\Python27\lib\urllib2.py", line 656, in http_error_302
return self.parent.open(new, timeout=req.timeout)
File "C:\Python27\lib\urllib2.py", line 437, in open
response = meth(req, response)
File "C:\Python27\lib\urllib2.py", line 550, in http_response
'http', request, response, code, msg, hdrs)
File "C:\Python27\lib\urllib2.py", line 475, in error
return self._call_chain(*args)
File "C:\Python27\lib\urllib2.py", line 409, in _call_chain
result = func(*args)
File "C:\Python27\lib\urllib2.py", line 558, in http_error_default
raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 401: authenticationrequired
But if I opened the page or other web page, and manually enter my credential, this script works fine after that. Am I missing something?
Just to add, my current network are using McAfee web gateway device. So sometimes we need to enter our credential to proceed browsing the net. Our user/pass are integrated with Active Directory. Is that may cause the issue?
This seems to work really well (taken from another thread)
import urllib2
import base64
import sys
user = 'myuserID'
password = "mypassword"
ip = sys.argv[1]
url = "http://www.websites.com/" + ip
request = urllib2.Request(url)
base64string = base64.encodestring('%s:%s' % (user, password)).replace('\n', '')
request.add_header("Authorization", "Basic %s" % base64string)
result = urllib2.urlopen(request)
Or you may use requests:
from requests.auth import HTTPBasicAuth
user = 'myuserID'
password = "mypassword"
ip = sys.argv[1]
url = "http://www.websites.com/" + ip
res=requests.get(url , auth=HTTPBasicAuth(user, password))
print res.text
I looked at this answer: Again urllib.error.HTTPError: HTTP Error 400: Bad Request because it was very similar to my question, but that solution did not work. I'm using Python 3.3.2. The lines that are like ..something.. are just values I replaced to protect my privacy. They should be strings
I'm getting an Error 400: Bad request from this code:
import urllib.parse
import urllib.request
url = '..url..'
user_agent = 'Mozilla/4.0 (compatible; MSIE 5.5; Windows NT)'
values = {'email':'..my email address..',
'github':'..my github account..'}
headers = {'User-Agent':user_agent}
data = urllib.parse.urlencode(values)
data = data.encode('utf-8')
req = urllib.request.Request(url, data, headers)
response = urllib.request.urlopen(req) #this line causes the errors
page = response.read()
This is the particular error message:
Traceback (most recent call last):
File "/Users/.../Documents/Code 2040/File.py", line 23, in <module>
response = urllib.request.urlopen(req)
File "/Library/Frameworks/Python.framework/Versions/3.3/lib/python3.3/urllib/request.py", line 156, in urlopen
return opener.open(url, data, timeout)
File "/Library/Frameworks/Python.framework/Versions/3.3/lib/python3.3/urllib/request.py", line 475, in open
response = meth(req, response)
File "/Library/Frameworks/Python.framework/Versions/3.3/lib/python3.3/urllib/request.py", line 587, in http_response
'http', request, response, code, msg, hdrs)
File "/Library/Frameworks/Python.framework/Versions/3.3/lib/python3.3/urllib/request.py", line 513, in error
return self._call_chain(*args)
File "/Library/Frameworks/Python.framework/Versions/3.3/lib/python3.3/urllib/request.py", line 447, in _call_chain
result = func(*args)
File "/Library/Frameworks/Python.framework/Versions/3.3/lib/python3.3/urllib/request.py", line 595, in http_error_default
raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 400: Bad Request
I'm trying to do this on remote Ubuntu Server:
>>> import urllib2, requests
>>> url = 'http://python.org/'
>>> urllib2.urlopen(url)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/lib/python2.7/urllib2.py", line 126, in urlopen
return _opener.open(url, data, timeout)
File "/usr/lib/python2.7/urllib2.py", line 406, in open
response = meth(req, response)
File "/usr/lib/python2.7/urllib2.py", line 519, in http_response
'http', request, response, code, msg, hdrs)
File "/usr/lib/python2.7/urllib2.py", line 444, in error
return self._call_chain(*args)
File "/usr/lib/python2.7/urllib2.py", line 378, in _call_chain
result = func(*args)
File "/usr/lib/python2.7/urllib2.py", line 527, in http_error_default
raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 404: Not Found
>>> requests.get(url)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/home/django/zyq2/venv/local/lib/python2.7/site-packages/requests/api.py", line 55, in get
return request('get', url, **kwargs)
File "/home/django/zyq2/venv/local/lib/python2.7/site-packages/requests/api.py", line 44, in request
return session.request(method=method, url=url, **kwargs)
File "/home/django/zyq2/venv/local/lib/python2.7/site-packages/requests/sessions.py", line 382, in request
resp = self.send(prep, **send_kwargs)
File "/home/django/zyq2/venv/local/lib/python2.7/site-packages/requests/sessions.py", line 505, in send
history = [resp for resp in gen] if allow_redirects else []
File "/home/django/zyq2/venv/local/lib/python2.7/site-packages/requests/sessions.py", line 99, in resolve_redir ts
raise TooManyRedirects('Exceeded %s redirects.' % self.max_redirects)
requests.exceptions.TooManyRedirects: Exceeded 30 redirects.
But it works fine on local Windows machine:
>>> urllib2.urlopen(url)
<addinfourl at 57470168 whose fp = <socket._fileobject object at 0x036CB630>>
>>> requests.get(url)
<Response [200]>
I have absolutely no idea about what's going on and would appreciate any suggestion.
Update
I tried S.M. Al Mamun's suggestion and got an exception with long traceback:
>>> req = urllib2.Request(url, headers={ 'User-Agent': 'Mozilla/5.0' })
>>> urllib2.urlopen(req).read()
...
long traceback (more than one page)
...
urllib2.HTTPError: HTTP Error 303: The HTTP server returned a redirect error that would lead to an infinite loop.
The last 30x error message was:
See Other
Infinite loop again (I mean TooManyRedirects exception).
Try using a user-agent:
req = urllib2.Request(url, headers={ 'User-Agent': 'Mozilla/5.0' })
urllib2.urlopen(req).read()
If it doesn't work still, that might be your Ubuntu is offline!
Hello Guys! I want to access some web page through python script. The url is: http://www.idealo.de/preisvergleich/Shop/27039.html
When I access it through web browser it is OK. But when I want to access it with urllib2:
a = urllib2.urlopen("http://www.idealo.de/preisvergleich/Shop/27039.html")
It gives me the following error:
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/lib/python2.7/urllib2.py", line 126, in urlopen
return _opener.open(url, data, timeout)
File "/usr/lib/python2.7/urllib2.py", line 406, in open
response = meth(req, response)
File "/usr/lib/python2.7/urllib2.py", line 519, in http_response
'http', request, response, code, msg, hdrs)
File "/usr/lib/python2.7/urllib2.py", line 444, in error
return self._call_chain(*args)
File "/usr/lib/python2.7/urllib2.py", line 378, in _call_chain
result = func(*args)
File "/usr/lib/python2.7/urllib2.py", line 527, in http_error_default
raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 403: Forbidden
Also I tried to access it with wget:
wget http://www.idealo.de/preisvergleich/Shop/27039.html
The error is:
--2012-04-23 12:42:03-- http://www.idealo.de/preisvergleich/Shop/27039.html
Resolving www.idealo.de (www.idealo.de)... 62.146.49.133
Connecting to www.idealo.de (www.idealo.de)|62.146.49.133|:80... connected.
HTTP request sent, awaiting response... 403 Forbidden
2012-04-23 12:42:03 ERROR 403: Forbidden.
Can anyone explain why it is so? And how can I access it using python?
They're blocking some user agents. If you try with the following:
wget -U "Mozilla/5.0" http://www.idealo.de/preisvergleich/Shop/27039.html
it works. So you have to find the way to fake the user agent in your python code to make it work.
Try this:
import urllib2
opener = urllib2.build_opener()
opener.addheaders = [('User-agent', 'Mozilla/5.0')]
a = opener.open("http://www.idealo.de/preisvergleich/Shop/27039.html")