问题
I followed these instructions to make a local wsdl. Some lines in my local WSDL:
< wsdl:definitions
xmlns:wsdl="http://schemas.xmlsoap.org/wsdl/"
xmlns:soap="http://schemas.xmlsoap.org/wsdl/soap/"
xmlns:soapenc="http://schemas.xmlsoap.org/soap/encoding/"
xmlns:wsu="http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-wssecurity-utility-1.0.xsd"
xmlns:xsd="http://www.w3.org/2001/XMLSchema"
xmlns:soap12="http://schemas.xmlsoap.org/wsdl/soap12/"
xmlns:tns="http://tempuri.org/"
xmlns:wsa="http://schemas.xmlsoap.org/ws/2004/08/addressing"
xmlns:wsp="http://schemas.xmlsoap.org/ws/2004/09/policy"
xmlns:i0="http://Microsoft.ServiceModel.Samples"
xmlns:wsap="http://schemas.xmlsoap.org/ws/2004/08/addressing/policy"
xmlns:wsaw="http://www.w3.org/2006/05/addressing/wsdl"
xmlns:msc="http://schemas.microsoft.com/ws/2005/12/wsdl/contract"
xmlns:wsa10="http://www.w3.org/2005/08/addressing"
xmlns:wsx="http://schemas.xmlsoap.org/ws/2004/09/mex"
xmlns:wsam="http://www.w3.org/2007/05/addressing/metadata"
name="CalculatorService"
targetNamespace="http://tempuri.org/">
When I use python like this:
from SOAPpy import WSDL
server = WSDL.Proxy('http://114.212.190.58:8000/ServiceModelSamples/service?wsdl')
server.ADD(a=3,b=4)
I get the error:
xml.parsers.expat.ExpatError: not well-formed(invalid token): line 1, column5.
回答1:
when I google basichttpbinding and wshttpbinding,I got Difference between BasicHttpBinding and WsHttpBinding
when I finished reading this article then I found a solution.
In How to: Host a WCF Service in a Managed Windows Service,the thirteenth step.you can replace wshttpBinding with basicHttpBinding in XMLlang file.
Finally everything goes well
Sorry for my poor English.
来源：https://stackoverflow.com/questions/15548002/when-i-consume-wsdl-using-python-i-get-an-xml-parsers-expat-expaterror