使用urllib发送XML

Question

在此链接之后，我尝试使用GET将XML文件发送到我的web服务:

import urllib
from createfile import XML

URL = "http://http://localhost:8080/mywebservice
parameter = urllib.urlencode({'XML': XML})
response = urllib.urlopen(URL + "?%s" % parameter)
print response.read()

但是它给了我这个错误:

Traceback (most recent call last):
  File "C:eclipsetesting_workspacehttp testersrcMain.py", line 15, in <module>
    response = urllib.urlopen(URL + "?%s" % parameter)
  File "C:Python27liburllib.py", line 84, in urlopen
    return opener.open(url)
  File "C:Python27liburllib.py", line 205, in open
    return getattr(self, name)(url)
  File "C:Python27liburllib.py", line 331, in open_http
    h = httplib.HTTP(host)
  File "C:Python27libhttplib.py", line 1047, in __init__
    self._setup(self._connection_class(host, port, strict))
  File "C:Python27libhttplib.py", line 681, in __init__
    self._set_hostport(host, port)
  File "C:Python27libhttplib.py", line 706, in _set_hostport
    raise InvalidURL("nonnumeric port: '%s'" % host[i+1:])
httplib.InvalidURL: nonnumeric port: ''

但是如果我使用该链接中描述的POST方法，它工作得很好，我的问题是我需要使用GET，那么为什么我得到这些错误?

response = urllib.urlopen(URL, parameter)  // this works

Answer 1

通过GET请求发送XML文件是完全没有意义的。

用POST代替