Urllib HTTP 错误 403

Question

我搜索了论坛，试图找出以下代码不起作用的原因：

import nltk, re, pprint
from urllib import request
url = "http://www.gutenberg.org/files/2554/2554.txt"
response = request.urlopen(url)
raw = response.read().decode('utf8')
print(raw[:75])

但迄今为止未能成功解决问题。以下是我尝试实施的一些类似解决方案，但无济于事： Forum 1, Forum 2

我得到的错误是：

 File "C:\Python33\lib\urllib\request.py", line 163, in urlopen
return opener.open(url, data, timeout)
 File "C:\Python33\lib\urllib\request.py", line 472, in open
response = meth(req, response)
 File "C:\Python33\lib\urllib\request.py", line 582, in http_response
'http', request, response, code, msg, hdrs)
 File "C:\Python33\lib\urllib\request.py", line 510, in error
return self._call_chain(*args)
 File "C:\Python33\lib\urllib\request.py", line 444, in _call_chain
result = func(*args)
 File "C:\Python33\lib\urllib\request.py", line 590, in http_error_default
raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 403: Forbidden

任何帮助将不胜感激

Answer 1

此代码有效：

Python 2

from  urllib import urlopen

url = "http://www.gutenberg.org/files/2554/2554.txt"
response = urlopen(url)

if response.code == 200:
    raw = response.read().decode('utf-8')
    print raw[:75]
else:
    print 'Error', response.code

response.close()

回复：

古腾堡计划罪与罚电子书，费奥多尔·陀思妥耶夫斯基（Fyodor Dostoevsky）

Python 3

from  urllib import request

url = "http://www.gutenberg.org/files/2554/2554.txt"

try:
    response = request.urlopen(url)
    raw = response.read().decode('utf-8')
    print(raw[:75])
except Exception as ex:
    print('Error:', ex)

如果你得到 HTTP 代码 403，这意味着你禁止访问这个 url。