Sending XML to a WEB Service and Getting Response Back

[John Nagle]

On 12/21/2010 11:26 AM, Anurag Chourasia wrote:
> Thanks for the response all.
>
> I tried exploring suds (which seems to be the current) and i hit
> problems right away. I will now try urllib or httplib.
>
> I have asked for help in the suds forum. Hope somebody replies.
>
> When i try to create a client, the error is as follows.
>
>>>> from suds.client import Client
>>>> url = 'http://10.251.4.33:8041/DteEnLinea/ws/EnvioGuia.jws'
>>>> client = Client(url)
>
> Traceback (most recent call last):
>    File "<stdin>", line 1, in<module>
>    File "suds/client.py", line 112, in __init__
>      self.wsdl = reader.open(url)
>    File "suds/reader.py", line 152, in open
>      d = self.fn(url, self.options)
>    File "suds/wsdl.py", line 136, in __init__
>      d = reader.open(url)
>    File "suds/reader.py", line 79, in open
>      d = self.download(url)
>    File "suds/reader.py", line 101, in download
>      return sax.parse(string=content)
>    File "suds/sax/parser.py", line 136, in parse
>      sax.parse(source)
>    File "/usr/local/lib/python2.7/xml/sax/expatreader.py", line 107, in parse
>      xmlreader.IncrementalParser.parse(self, source)
>    File "/usr/local/lib/python2.7/xml/sax/xmlreader.py", line 123, in parse
>      self.feed(buffer)
>    File "/usr/local/lib/python2.7/xml/sax/expatreader.py", line 211, in feed
>      self._err_handler.fatalError(exc)
>    File "/usr/local/lib/python2.7/xml/sax/handler.py", line 38, in fatalError
>      raise exception
> xml.sax._exceptions.SAXParseException:<unknown>:1:62: syntax error
>>>> [3] + Stopped (SIGTSTP)        python
>
> This seems to be a old problem passing versions.
>
> Regards,
> Anurag

    Try posting a URL that isn't on network 10. That's some local
network at your end.

					John Nagle