urllib2 HTTP错误400：错误的请求

Question

I have a piece of code like this

host = 'http://www.bing.com/search?q=%s&go=&qs=n&sk=&sc=8-13&first=%s' % (query, page)
req = urllib2.Request(host)
req.add_header('User-Agent', User_Agent)
response = urllib2.urlopen(req)

and when I input a query greater than one word like "the dog" i get the following error.

response = urllib2.urlopen(req)
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 400, in open
response = meth(req, response)
File "/usr/lib/python2.7/urllib2.py", line 513, in http_response
'http', request, response, code, msg, hdrs)
File "/usr/lib/python2.7/urllib2.py", line 438, in error
return self._call_chain(*args)
File "/usr/lib/python2.7/urllib2.py", line 372, in _call_chain
result = func(*args)
File "/usr/lib/python2.7/urllib2.py", line 521, in http_error_default
raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 400: Bad Request

Can anyone point out what im doing wrong? Thanks in advance.

Answer 1

The reason that "the dog" returns a 400 Error is because you aren't escaping the string for a URL.

If you do this:

import urllib, urllib2

quoted_query = urllib.quote(query)
host = 'http://www.bing.com/search?q=%s&go=&qs=n&sk=&sc=8-13&first=%s' % (quoted_query, page)
req = urllib2.Request(host)
req.add_header('User-Agent', User_Agent)
response = urllib2.urlopen(req)

It will work.

However I highly suggest you use requests instead of using urllib/urllib2/httplib. It's much much easier and it'll handle all of this for you.

This is the same code with python requests:

import requests

results = requests.get("http://www.bing.com/search", 
              params={'q': query, 'first': page}, 
              headers={'User-Agent': user_agent})

Answer 2

You need to use urllib.quote() on your 'query' variable:

query = urllib.quote(query)
host = 'http://www.bing.com/search?q=%s&go=&qs=n&sk=&sc=8-13&first=%s' % (query, page)

This does the necessary URL escaping to convert the space in big dog to big%20dog .

Answer 3

您必须使用urllib.quote

Answer 4

Here is an example of how to use urllib.request object in Python 3.6 and above.

import urllib.request
import json
from pprint import pprint

url = "some_url"

values = {
    "first_name": "Vlad",
    "last_name": "Bezden",
    "urls": [
        "https://twitter.com/VladBezden",
        "https://github.com/vlad-bezden",
    ],
}


headers = {
    "Content-Type": "application/json",
    "Accept": "application/json",
}

data = json.dumps(values).encode("utf-8")
pprint(data)

try:
    req = urllib.request.Request(url, data, headers)
    with urllib.request.urlopen(req) as f:
        res = f.read()
    pprint(res.decode())
except Exception as e:
    pprint(e)

Answer 5

I also encountered the same problem. Turns out the problem was the method was set inappropriately. When you include urlencoded data in urllib2.urlopen () the method should be set to POST and when you exclude it, method should be GET. So, how do you set the method is given below:

For POST request

request_object = urllib2.Request(url)
method = ("POST", "GET")
request_object.get_method = lambda: method[0] #If method is set to POST
url_handle = opener.open(req, data) #If method is set to POST

For GET request

request_object = urllib2.Request(url)
method = ("POST", "GET")
request_object.get_method = lambda: method[1] #If method is set to GET
url_handle = opener.open(req) #If method is set to GET

This will set your url request method to the appropriate required method