I'm using SOPApy to make a client for SOAP. Following code rises error.
import SOAPpy
wsdlFile = 'https://10.10.10.10/services/fwif?wsdl'
proxy = SOAPpy.WSDL.Proxy(wsdlFile)
Traceback (most recent call last):
File "run.py", line 28, in <module>
proxy = SOAPpy.WSDL.Proxy(wsdlFile)
File "/home/dinn/miniconda/envs/soaptest/lib/python2.7/site-packages/SOAPpy/WSDL.py", line 83, in __init__
self.wsdl = reader.loadFromString(str(wsdlsource))
File "/home/dinn/miniconda/envs/soaptest/lib/python2.7/site-packages/wstools/WSDLTools.py", line 49, in loadFromString
return self.loadFromStream(StringIO(data))
File "/home/dinn/miniconda/envs/soaptest/lib/python2.7/site-packages/wstools/WSDLTools.py", line 28, in loadFromStream
document = DOM.loadDocument(stream)
File "/home/dinn/miniconda/envs/soaptest/lib/python2.7/site-packages/wstools/Utility.py", line 645, in loadDocument
return xml.dom.minidom.parse(data)
File "/home/dinn/miniconda/envs/soaptest/lib/python2.7/xml/dom/minidom.py", line 1918, in parse
return expatbuilder.parse(file)
File "/home/dinn/miniconda/envs/soaptest/lib/python2.7/xml/dom/expatbuilder.py", line 928, in parse
result = builder.parseFile(file)
File "/home/dinn/miniconda/envs/soaptest/lib/python2.7/xml/dom/expatbuilder.py", line 207, in parseFile
parser.Parse(buffer, 0)
xml.parsers.expat.ExpatError: not well-formed (invalid token): line 1, column 6
On stackoverflow I found a similar problem When I consume wsdl using python, I get an xml.parsers.expat.ExpatError but don't understand how to apply it for SOAPpy.
