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I'm using the Python zeep library but the documentation seems a bit thin. I'd like to be able to specify the URL of the SOAP server but at the moment I can only do this by hacking the WSDL file.

fedorqui
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Nick Keighley
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1 Answers1

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I've just added this feature (https://github.com/mvantellingen/python-zeep/commit/08a757c35ad34defc401baea04b23ae889e2ee1a)

See the docs http://docs.python-zeep.org/en/latest/in-depth.html#overriding-the-default-endpoint-address

mvantellingen
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  • well I can't knock your speed of response! – Nick Keighley Jun 02 '16 at 10:11
  • I've testing it and and it fixes my problem. – Nick Keighley Jun 03 '16 at 09:33
  • Can you change your answer to give a longer example? I'm getting `TypeError: create_service() takes 3 positional arguments but 4 were given` when I try to fix a https service that incorrectly reports to be a http service. The command is `service = client.create_service(r' {http://localhost/}SuperServiceSoap',r'{http://localhost/}SuperServiceSoap12',r'https://info.removed.com:443/StatusInfo/')` - – 576i Aug 30 '16 at 14:54
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    This answer is no longer true because the `set_address` feature has been removed. see this answer instead: https://stackoverflow.com/questions/42236251/change-service-url-in-python-zeep – baldychristophe Oct 12 '17 at 15:59