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I am a newbie trying to write a script of a web spider. I want to go to a page, enter a data in a textbox, go to the next page by clicking the submit button and retrieve all data on the new page, iteratively.

The following is the code I am trying:

import urllib
import urllib2
import string
import sys
from BeautifulSoup import BeautifulSoup

hdr = {'User-Agent': 'Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.11 (KHTML, like Gecko) Chrome/23.0.1271.64 Safari/537.11','Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8','Accept-Charset': 'ISO-8859-1,utf-8;q=0.7,*;q=0.3','Accept-Encoding': 'none','Accept-Language': 'en-US,en;q=0.8','Connection': 'keep-alive'}
values = {'query' : '5ed10c844ed4266a18d34e2ba06b381a' }
data = urllib.urlencode(values)
request = urllib2.Request("https://www.virustotal.com/#search", data, headers=hdr)
response = urllib2.urlopen(request)
the_page = response.read()
pool = BeautifulSoup(the_page)

print pool

The following is the error:

Traceback (most recent call last):
File "C:\Users\Dipanshu\Desktop\webscraping_demo.py", line 19, in <module>
response = urllib2.urlopen(request)
File "C:\Python27\lib\urllib2.py", line 126, in urlopen
return _opener.open(url, data, timeout)
File "C:\Python27\lib\urllib2.py", line 406, in open
response = meth(req, response)
File "C:\Python27\lib\urllib2.py", line 519, in http_response
'http', request, response, code, msg, hdrs)
File "C:\Python27\lib\urllib2.py", line 444, in error
return self._call_chain(*args)
File "C:\Python27\lib\urllib2.py", line 378, in _call_chain
result = func(*args)
File "C:\Python27\lib\urllib2.py", line 527, in http_error_default
raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
HTTPError: HTTP Error 403: Forbidden

How can I solve this?

4

2 回答 2

1 
from bs4 import BeautifulSoup
import urllib.request

user_agent = 'Mozilla/5.0'
headers = {'User-Agent': user_agent }
target_url = 'https://www.google.co.kr/search?q=cat&source=lnms&tbm=isch&sa=X&ved=0ahUKEwjtrZCg7uXbAhVaUd4KHc2HDgIQ_AUICygC&biw=1375&bih=842'

request = urllib.request.Request( url=target_url, headers=headers )
req = urllib.request.urlopen(request)
soup = BeautifulSoup(req.read(), 'html.parser')

target_url ：谷歌搜索网页“猫”

“标题”将帮助您完成 Forbidden 错误。这段代码

于 2018-06-21T23:07:07.890 回答
0

据我了解，您的request参数设置不正确，并且（可能）将您的蜘蛛驱动到您不应该查看的页面。

该用户遇到了类似的问题，但通过修改标题修复了它。

于 2012-12-21T09:42:14.927 回答