Question

我有一个我正在测试的简单网站。它在localhost上运行，我可以在我的Web浏览器中访问它。索引页面只是“运行”一词。 urllib.urlopen会成功阅读该页面，但urllib2.urlopen不会。这是一个演示问题的脚本（这是实际的脚本而不是简化不同的测试脚本）：

import urllib, urllib2
print urllib.urlopen("http://127.0.0.1").read()  # prints "running"
print urllib2.urlopen("http://127.0.0.1").read() # throws an exception

这是堆栈跟踪：

Traceback (most recent call last):
  File "urltest.py", line 5, in <module>
    print urllib2.urlopen("http://127.0.0.1").read()
  File "C:\Python25\lib\urllib2.py", line 121, in urlopen
    return _opener.open(url, data)
  File "C:\Python25\lib\urllib2.py", line 380, in open
    response = meth(req, response)
  File "C:\Python25\lib\urllib2.py", line 491, in http_response
    'http', request, response, code, msg, hdrs)
  File "C:\Python25\lib\urllib2.py", line 412, in error
    result = self._call_chain(*args)
  File "C:\Python25\lib\urllib2.py", line 353, in _call_chain
    result = func(*args)
  File "C:\Python25\lib\urllib2.py", line 575, in http_error_302
    return self.parent.open(new)
  File "C:\Python25\lib\urllib2.py", line 380, in open
    response = meth(req, response)
  File "C:\Python25\lib\urllib2.py", line 491, in http_response
    'http', request, response, code, msg, hdrs)
  File "C:\Python25\lib\urllib2.py", line 418, in error
    return self._call_chain(*args)
  File "C:\Python25\lib\urllib2.py", line 353, in _call_chain
    result = func(*args)
  File "C:\Python25\lib\urllib2.py", line 499, in http_error_default
    raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 504: Gateway Timeout

有什么想法吗？我可能最终需要urllib2的一些更高级的功能，所以我不想只使用urllib，而且我想了解这个问题。

Answer 1

听起来你已经定义了urllib2正在接收的代理设置。当它尝试代理“127.0.0.01/”时，代理放弃并返回504错误。

来自Obscure python urllib2 proxy gotcha：

proxy_support = urllib2.ProxyHandler({})
opener = urllib2.build_opener(proxy_support)
print opener.open("http://127.0.0.1").read()

# Optional - makes this opener default for urlopen etc.
urllib2.install_opener(opener)
print urllib2.urlopen("http://127.0.0.1").read()

Answer 2

首先调用urlib2.open后跟urllib.open会有相同的结果吗？只是想知道第一次打开是否导致http服务器忙于导致超时？

Answer 3

我不知道发生了什么，但你可能会发现这有助于搞清楚：

>>> import urllib2
>>> urllib2.urlopen('http://mit.edu').read()[:10]
'<!DOCTYPE '
>>> urllib2._opener.handlers[1].set_http_debuglevel(100)
>>> urllib2.urlopen('http://mit.edu').read()[:10]
connect: (mit.edu, 80)
send: 'GET / HTTP/1.1\r\nAccept-Encoding: identity\r\nHost: mit.edu\r\nConnection: close\r\nUser-Agent: Python-urllib/2.5\r\n\r\n'
reply: 'HTTP/1.1 200 OK\r\n'
header: Date: Tue, 14 Oct 2008 15:52:03 GMT
header: Server: MIT Web Server Apache/1.3.26 Mark/1.5 (Unix) mod_ssl/2.8.9 OpenSSL/0.9.7c
header: Last-Modified: Tue, 14 Oct 2008 04:02:15 GMT
header: ETag: "71d3f96-2895-48f419c7"
header: Accept-Ranges: bytes
header: Content-Length: 10389
header: Connection: close
header: Content-Type: text/html
'<!DOCTYPE '

Answer 4

urllib.urlopen（）在服务器上抛出以下请求：

GET / HTTP/1.0
Host: 127.0.0.1
User-Agent: Python-urllib/1.17

而urllib2.urlopen（）抛出这个：

GET / HTTP/1.1
Accept-Encoding: identity
Host: 127.0.0.1
Connection: close
User-Agent: Python-urllib/2.5

因此，您的服务器要么不了解HTTP / 1.1或额外的标头字段。

urllib.urlopen工作，但urllib2.urlopen没有

4 个答案: