HTTP Error 406: Not Acceptable Python urllib2 - python

I get the following error with the code below.
HTTP Error 406: Not Acceptable Python urllib2
This is my first step before I use beautifulsoup to parse the page.
import urllib2
opener = urllib2.build_opener()
opener.addheaders = [('User-agent', 'Mozilla/5.0')]
url = "http://www.choicemoney.us/retail.php"
response = opener.open(url)
All help greatly appreciated.


The resource identified by the request is only capable of generating
response entities which have content characteristics not acceptable
according to the accept headers sent in the request. [RFC2616]
Based on the code and what the RFC describes I assume that you need to set both the key and the value of the User-Agent header correctly.
These are correct examples:
Mozilla/5.0 (X11; U; Linux i686) Gecko/20071127 Firefox/2.0.0.11
Mozilla/5.0 (Windows NT 6.1) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/41.0.2228.0 Safari/537.36
Mozilla/5.0 (Macintosh; Intel Mac OS X 10_9_3) AppleWebKit/537.75.14 (KHTML, like Gecko) Version/7.0.3 Safari/7046A194A
Just replace the following.
opener.addheaders = [('User-agent', 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_9_3) AppleWebKit/537.75.14 (KHTML, like Gecko) Version/7.0.3 Safari/7046A194A')]


I believe #ipinak's answer is correct.
urllib2 actually provides a default User-Agent that works here, so if you delete opener.addheaders = [('User-agent', 'Mozilla/5.0')] the response should have status code 200.
I recommend the popular requests library for such jobs as its API is much easier to use.
url = "http://www.choicemoney.us/retail.php"
resp = requests.get(url)
print resp.status_code # 200
print resp.content # can be used in your beautifulsoup.

Related

Urllib request takes too long to respond [duplicate]

I try to open url with python3:
import urllib.request
fp = urllib.request.urlopen("http://lebed.com/")
mybytes = fp.read()
mystr = mybytes.decode("utf8")
fp.close()
print(mystr)
But it hangs on second line.
What's the reason of this problem and how to fix it?

I suppose the reason is that the url does not support robot visiting a site visit. You need to fake a browser visit by sending browser headers along with your request
import urllib.request
url = "http://lebed.com/"
req = urllib.request.Request(
url,
data=None,
headers={
'User-Agent': 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_9_3) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/35.0.1916.47 Safari/537.36'
}
)
f = urllib.request.urlopen(req)
Tried this one on my system and it works.

Agree with Arpit Solanki. Shown output for a failed request vs successful.
Failed
GET / HTTP/1.1
Accept-Encoding: identity
Host: www.lebed.com
Connection: close
User-Agent: Python-urllib/3.5
Success
GET / HTTP/1.1
Accept-Encoding: identity
Host: www.lebed.com
Connection: close
User-Agent: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_9_3) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/35.0.1916.47 Safari/537.36

Python Requests Get not Working

I have a simple Get request I'd like to make using Python's Request library.
import requests
HEADERS = {'user-agent': ('Mozilla/5.0 (Macintosh; Intel Mac OS X 10_10_5)'
'AppleWebKit/537.36 (KHTML, like Gecko)'
'Chrome/45.0.2454.101 Safari/537.36'),
'referer': 'http://stats.nba.com/scores/'}
url = 'http://stats.nba.com/stats/playbyplayv2?EndPeriod=10&EndRange=55800&GameID=0021500281&RangeType=2&Season=2016-17&SeasonType=Regular+Season&StartPeriod=1&StartRange=0'
response = requests.get(url, timeout=5, headers=HEADERS)
However, when I make the requests.get call, I get the error requests.exceptions.ReadTimeout: HTTPConnectionPool(host='stats.nba.com', port=80): Read timed out. (read timeout=5). But I am able to copy/paste that url into my browser and view the resulting JSON. Why is requests not able to get the result?

Your HEADERS format is wrong. I tried with this code and it worked without any issues:
import requests
HEADERS = {
'User-Agent': 'Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/67.0.3396.87 Safari/537.36',
}
url = 'http://stats.nba.com/stats/playbyplayv2?EndPeriod=10&EndRange=55800&GameID=0021500281&RangeType=2&Season=2016-17&SeasonType=Regular+Season&StartPeriod=1&StartRange=0'
response = requests.get(url, timeout=5, headers=HEADERS)
print(response.text)

Able to see image on browser, but urllib.urlretrieve() fails to downlad it. How can I download it?

Image path --> http://markinternational.info/data/out/366/221983609-black-hd-desktop-wallpaper.jpg
Code I am using
import urllib
urllib.urlretrieve("https://markinternational.info/data/out/366/221983609-black-hd-desktop-wallpaper.jpg" , "photu.jpg")
What it returns (returns same thing for successful or unsuccessful attempts)
('photu.jpg', <httplib.HTTPMessage instance at 0x7fe3cfb27d88>)
Can someone help?

You need to fake the user-agent to bypass this restriction by the web server.
I used Python3 and requests library, I managed to get the picture:
import requests
headers = {'User-Agent': 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_10_1) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/39.0.2171.95 Safari/537.36'}
url = 'https://markinternational.info/data/out/366/221983609-black-hd-desktop-wallpaper.jpg'
res = requests.get(url, headers=headers)
with open('photo.jpg', 'wb') as W:
W.write(res.content)

This might help.
import urllib
f = open('photu.jpg','wb')
f.write(urllib.urlopen('https://markinternational.info/data/out/366/221983609-black-hd-desktop-wallpaper.jpg').read())
f.close()

Since you're sending a raw http request without any User-Agent header, the server is not allowing the request to pass through. You can mock it with a defined User-Agent in header and it'll work as if it works on browser.
url = "https://markinternational.info/data/out/366/221983609-black-hd-desktop-wallpaper.jpg"
req = urllib.request.Request(
url,
data=None,
headers={
'User-Agent': 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_9_3) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/35.0.1916.47 Safari/537.36'
}
)
with open('image.jpg', 'wb') as img_file:
img_file.write(urllib.request.urlopen(req).read())

python user-agent xpath amazon

I am trying to make a scan of the HTML in 2 requests.
at the first one, it's working but when I am trying to use another one,
The HTML I am trying to locate are not visible its getting it wrong.
headers = {'User-Agent': 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_10_1) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/39.0.2171.95 Safari/537.36'}
page = requests.get('https://www.amazon.com/gp/aw/ol/B00DZKQSRQ/ref=mw_dp_olp?ie=UTF8&condition=new, headers=headers)
newHeader = {'User-Agent': 'Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/52.0.2743.82 Safari/537.36'}
pagePrice = requests.get('https://www.amazon.com/gp/aw/ol/B01EQJU8AW/ref=mw_dp_olp?ie=UTF8&condition=new',headers=newHeader)
The first request works fine and gets me the good HTML.
The second request gives bad HTML.
I saw this package, but not success :
https://pypi.python.org/pypi/fake-useragent
And I saw this topic not unaswered :
Double user-agent tag, "user-agent: user-agent: Mozilla/"
Thank you very much!

Changing user agent on urllib2.urlopen

How can I download a webpage with a user agent other than the default one on urllib2.urlopen?

I answered a similar question a couple weeks ago.
There is example code in that question, but basically you can do something like this: (Note the capitalization of User-Agent as of RFC 2616, section 14.43.)
opener = urllib2.build_opener()
opener.addheaders = [('User-Agent', 'Mozilla/5.0')]
response = opener.open('http://www.stackoverflow.com')

headers = { 'User-Agent' : 'Mozilla/5.0' }
req = urllib2.Request('www.example.com', None, headers)
html = urllib2.urlopen(req).read()
Or, a bit shorter:
req = urllib2.Request('www.example.com', headers={ 'User-Agent': 'Mozilla/5.0' })
html = urllib2.urlopen(req).read()

Setting the User-Agent from everyone's favorite Dive Into Python.
The short story: You can use Request.add_header to do this.
You can also pass the headers as a dictionary when creating the Request itself, as the docs note:
headers should be a dictionary, and will be treated as if add_header() was called with each key and value as arguments. This is often used to “spoof” the User-Agent header, which is used by a browser to identify itself – some HTTP servers only allow requests coming from common browsers as opposed to scripts. For example, Mozilla Firefox may identify itself as "Mozilla/5.0 (X11; U; Linux i686) Gecko/20071127 Firefox/2.0.0.11", while urllib2‘s default user agent string is "Python-urllib/2.6" (on Python 2.6).

For python 3, urllib is split into 3 modules...
import urllib.request
req = urllib.request.Request(url="http://localhost/", headers={'User-Agent':' Mozilla/5.0 (Windows NT 6.1; WOW64; rv:12.0) Gecko/20100101 Firefox/12.0'})
handler = urllib.request.urlopen(req)

All these should work in theory, but (with Python 2.7.2 on Windows at least) any time you send a custom User-agent header, urllib2 doesn't send that header. If you don't try to send a User-agent header, it sends the default Python / urllib2
None of these methods seem to work for adding User-agent but they work for other headers:
opener = urllib2.build_opener(proxy)
opener.addheaders = {'User-agent':'Custom user agent'}
urllib2.install_opener(opener)
request = urllib2.Request(url, headers={'User-agent':'Custom user agent'})
request.headers['User-agent'] = 'Custom user agent'
request.add_header('User-agent', 'Custom user agent')

For urllib you can use:
from urllib import FancyURLopener
class MyOpener(FancyURLopener, object):
version = 'Mozilla/5.0 (Windows; U; Windows NT 5.1; it; rv:1.8.1.11) Gecko/20071127 Firefox/2.0.0.11'
myopener = MyOpener()
myopener.retrieve('https://www.google.com/search?q=test', 'useragent.html')

Another solution in urllib2 and Python 2.7:
req = urllib2.Request('http://www.example.com/')
req.add_unredirected_header('User-Agent', 'Custom User-Agent')
urllib2.urlopen(req)

there are two properties of urllib.URLopener() namely:
addheaders = [('User-Agent', 'Python-urllib/1.17'), ('Accept', '*/*')] and
version = 'Python-urllib/1.17'.
To fool the website you need to changes both of these values to an accepted User-Agent. for e.g.
Chrome browser : 'Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/33.0.1750.149 Safari/537.36'
Google bot : 'Googlebot/2.1'
like this
import urllib
page_extractor=urllib.URLopener()
page_extractor.addheaders = [('User-Agent', 'Googlebot/2.1'), ('Accept', '*/*')]
page_extractor.version = 'Googlebot/2.1'
page_extractor.retrieve(<url>, <file_path>)
changing just one property does not work because the website marks it as a suspicious request.

Try this :
html_source_code = requests.get("http://www.example.com/",
headers={'User-Agent':'Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/44.0.2403.107 Safari/537.36',
'Upgrade-Insecure-Requests': '1',
'x-runtime': '148ms'},
allow_redirects=True).content

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