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我正在使用 SOPApy 为 SOAP 创建客户端。以下代码出现错误。

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

在stackoverflow上我发现了一个类似的问题当我使用python使用wsdl时，我得到一个xml.parsers.expat.ExpatError，但不明白如何将它应用于SOAPpy。

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0 回答 0