On Thu, 31 Jan 2008 11:03:26 -0800, Bernard wrote: > Hey y'all, > > Is there a way to POST a handmade SOAP request *without* using any > libraries like SOAPpy?
Yes, it's quite easy to SOAP by hand. I use Oren Tirosh's ElementBuilder class (on top of lxml instead of ElementTree) to build the SOAP request and the xpath capabilities in lxml to pull out the data I need from the response. http://www.tothink.com/python/ElementBuilder/ http://codespeak.net/lxml/ An incomplete example for contructing a request looks something like this: body = Element('soap:Envelope', { 'xmlns:soap': nss['soap']}, Element('soap:Header'), Element('soap:Body', { 'xmlns:msgs': nss['msgs'] }, Element('msgs:login', Element('msgs:passport', { 'xmlns:core': nss['core'] }, Element('core:password', password), Element('core:account', account))))) I use httplib2 for sending the HTTP requests: http://code.google.com/p/httplib2/ Incomplete example: headers['SOAPAction'] = action headers['Content-length'] = str(len(etree.tostring(body))) response, content = self._client.request( self.ns_uri, "POST", body=etree.tostring(body), headers=self._headers) if response.status == 500 and not \ (response["content-type"].startswith("text/xml") and \ len(content) > 0): raise HTTPError(response.status, content) if response.status not in (200, 500): raise HTTPError(response.status, content) doc = etree.parse(StringIO(content)) if response.status == 500: faultstring = doc.findtext(".//faultstring") raise HTTPError(response.status, faultstring) Now it's just a matter of using xpath expressions to dig into the "doc" structure for the bits you need. -- http://mail.python.org/mailman/listinfo/python-list
