Re: sending a handmade SOAP request

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.
.

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