Question

我正在使用BeautifulSoup和urlopen进行webscrape。当我尝试使用urlopen打开url时，提示出现大错误。我在做什么错了？

我使用python 3.7.2，我已经安装了Bs4和urlopen。

import bs4
from urllib.request import urlopen
from bs4 import BeautifulSoup as soup
my_url = 'https://ebgames.com.au/search?q=Skyrim'
var = urlopen(my_url)

ERROR:

Traceback (most recent call last):
  File "<pyshell#8>", line 1, in <module>
    urlopen (my_url)
  File "C:\Users\%USERNAME%\AppData\Local\Programs\Python\Python37\lib\urllib\request.py", line 222, in urlopen
    return opener.open(url, data, timeout)
  File "C:\Users\%USERNAME%\AppData\Local\Programs\Python\Python37\lib\urllib\request.py", line 531, in open
    response = meth(req, response)
  File "C:\Users\%USERNAME%\AppData\Local\Programs\Python\Python37\lib\urllib\request.py", line 641, in http_response
    'http', request, response, code, msg, hdrs)
  File "C:\Users\%USERNAME%\AppData\Local\Programs\Python\Python37\lib\urllib\request.py", line 569, in error
    return self._call_chain(*args)
  File "C:\Users\%USERNAME%\AppData\Local\Programs\Python\Python37\lib\urllib\request.py", line 503, in _call_chain
    result = func(*args)
  File "C:\Users\%USERNAME%\AppData\Local\Programs\Python\Python37\lib\urllib\request.py", line 649, in http_error_default
    raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 403: Forbidden

预期结果没有错误。

Answer 1

尝试将浏览器用户代理设置为您的请求作为请求标头：

import bs4
from urllib.request import Request, urlopen
from bs4 import BeautifulSoup as soup
my_url = req = Request('https://ebgames.com.au/search?q=Skyrim', headers={'User-Agent': 'Mozilla/5.0'})
var = urlopen(my_url).read()

有关更多信息，请参阅此answer

将变量分配给urlopen（my_url）时出错

1 个答案: