Send XML data to this interface address as needed

Specifically, I have an interface address that sends XML data to this interface address as needed. Then, a response is received, and the response content is processed accordingly.
I have found a piece of code on the Internet, which should be what I need, but I don't quite understand some of it. I 'd like to ask you here. The Code is as follows. Please help me with it. Python code?

1234567891011121314151617181920212223242526272829

import urllib2 import sys, httplib def SendRtx(target,title,content):     SENDTPL = \             '''<?xml version="1.0" encoding="UTF-8"?>     <SOAP-ENV:Envelope xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/" xmlns:ns1="http://ws.oa.com/common/message">         <SOAP-ENV:Body>             <ns1:SendRTX>                 <ns1:sender>dantezhu</ns1:sender>                 <ns1:receiver>%s</ns1:receiver>                 <ns1:title>%s</ns1:title>                 <ns1:msgInfo>%s</ns1:msgInfo>                 <ns1:messageType>0</ns1:messageType>             </ns1:SendRTX>         </SOAP-ENV:Body>     </SOAP-ENV:Envelope>'''     SoapMessage = SENDTPL % (target,title,content)     webservice = httplib.HTTP("ws.oa.com")     webservice.putrequest("POST", "/messageservice.asmx")     webservice.putheader("Host", "ws.oa.com")     webservice.putheader("User-Agent", "Python Post")     webservice.putheader("Content-type", "text/xml; charset=\"UTF-8\"")     webservice.putheader("Content-length", "%d" % len(SoapMessage))     webservice.putheader("SOAPAction", "\"http://ws.oa.com/common/message/SendRTX\"")     webservice.endheaders()     webservice.send(SoapMessage)     # get the response     statuscode, statusmessage, header = webservice.getreply()