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我尝试使用 urllib2 发送带有 json 正文的 post 请求：

request = urllib2.Request('http://localhost:8090/api/', jdata, {'Content-Type': 'application/json'})
f = urllib2.urlopen(request)

但得到例外：

  File "/usr/lib/python2.7/urllib2.py", line 127, in urlopen
    return _opener.open(url, data, timeout)
  File "/usr/lib/python2.7/urllib2.py", line 407, in open
    response = meth(req, response)
  File "/usr/lib/python2.7/urllib2.py", line 520, in http_response
    'http', request, response, code, msg, hdrs)
  File "/usr/lib/python2.7/urllib2.py", line 439, in error
    result = self._call_chain(*args)
  File "/usr/lib/python2.7/urllib2.py", line 379, in _call_chain
    result = func(*args)
  File "/usr/lib/python2.7/urllib2.py", line 626, in http_error_302
    return self.parent.open(new, timeout=req.timeout)
  File "/usr/lib/python2.7/urllib2.py", line 401, in open
    response = self._open(req, data)
  File "/usr/lib/python2.7/urllib2.py", line 419, in _open
    '_open', req)
  File "/usr/lib/python2.7/urllib2.py", line 379, in _call_chain
    result = func(*args)
  File "/usr/lib/python2.7/urllib2.py", line 1211, in http_open
    return self.do_open(httplib.HTTPConnection, req)
  File "/usr/lib/python2.7/urllib2.py", line 1150, in do_open
    h = http_class(host, timeout=req.timeout) # will parse host:port
  File "/usr/lib/python2.7/httplib.py", line 693, in __init__
    self._set_hostport(host, port)
  File "/usr/lib/python2.7/httplib.py", line 721, in _set_hostport
    raise InvalidURL("nonnumeric port: '%s'" % host[i+1:])
httplib.InvalidURL: nonnumeric port: '8090��>��hbWb�ba'

我该如何解决？

谢谢你。

4

1 回答 1

0

似乎 urllib2 函数无法正确解析您的请求。你为什么不试试卷曲？

import pycurl

c = pycurl.Curl()
c.setopt(c.URL, 'http://myfavpizzaplace.com/order')
c.setopt(c.POSTFIELDS, 'pizza=Quattro+Stagioni&extra=cheese')
c.setopt(c.VERBOSE, True)
c.perform()

来自http://www.angryobjects.com/2011/10/15/http-with-python-pycurl-by-example/的示例

于 2013-04-27T14:49:22.947 回答