Python PySimpleSAP-运行时错误:没有为url提供方案
我第一次尝试使用pysimplesoap和我第一次尝试soap代码Python PySimpleSAP-运行时错误:没有为url提供方案,python,soap,xsd,wsdl,pysimplesoap,Python,Soap,Xsd,Wsdl,Pysimplesoap,我第一次尝试使用pysimplesoap和我第一次尝试soap代码 from pysimplesoap.client import SoapClient j_location = 'http://api.jasperwireless.com/ws/schema' j_xsd = 'http://api.jasperwireless.com/ws/schema/JasperAPI.xsd' j_echo_wsdl = 'http://api.jasperwireless.com/ws/schem
from pysimplesoap.client import SoapClient
j_location = 'http://api.jasperwireless.com/ws/schema'
j_xsd = 'http://api.jasperwireless.com/ws/schema/JasperAPI.xsd'
j_echo_wsdl = 'http://api.jasperwireless.com/ws/schema/Echo.wsdl'
j_billing_wsdl = 'http://api.jasperwireless.com/ws/schema/Billing.wsdl'
print 'Creating client'
myclient = SoapClient(wsdl=j_echo_wsdl)
print 'Target Namespace', myclient.namespace
错误
运行时错误:没有为url:JasperAPI.xsd提供方案
我不知道该如何解决这个错误 我想问题是因为JasperAPI.xsd在WSDL中被引用为本地文件:
<xs:import namespace="http://api.jasperwireless.com/ws/schema" schemaLocation="JasperAPI.xsd"/>
进入
但至少libxml2——大多数人都在使用它?Python SOAP实现—不这样做
作为一次性快速修复,您可以尝试将JasperAPI.xsd放在本地工作目录中
schemaLocation="JasperAPI.xsd"
schemaLocation="http://api.jasperwireless.com/ws/schema/JasperAPI.xsd"