Deploy multiple web services, i.e. multiple wsdl files, in python

问题:

I'm creating web services in python using Spyne based on this example. However, all my services are combined into one wsdl file locating at http://localhost:8000/?wsdl. I'm looking for another way to deploy each web service separately in a single wsdl file, e.g. http://localhost:8000/service1/?wsdl and http://localhost:8000/service2?wsdl

回答1:

Spyne has a WsgiMounter class for this:

from spyne.util.wsgi_wrapper import WsgiMounter  app1 = Application([SomeService], tns=tns,         in_protocol=Soap11(), out_protocol=Soap11()) app2 = Application([SomeOtherService], tns=tns,         in_protocol=Soap11(), out_protocol=Soap11()) wsgi_app = WsgiMounter({     'app1': app1,     'app2': app2, })

Now you can pass wsgi_app to the Wsgi implementation that you're using the same way you'd pass a WsgiApplication instance.

Your Wsgi implementation also would definitely have a similar functionality, you can also use that in case e.g. you need to serve something for the root request instead of an empty 404 request.

An up-to-date fully working example can be found at: https://github.com/plq/spyne/blob/master/examples/multiple_protocols/server.py

Please note that you can't use one Service class with multiple applications. If you must do that, you can do it like this:

def SomeServiceFactory():     class SomeService(ServiceBase):         @rpc(Unicode, _returns=Unicode)         def echo_string(ctx, string):             return string     return SomeService

and use the SomeServiceFactory() call for every Application instance.

e.g.

app1 = Application([SomeServiceFactory()], tns=tns,         in_protocol=Soap11(), out_protocol=Soap11()) app2 = Application([SomeServiceFactory()], tns=tns,         in_protocol=Soap11(), out_protocol=Soap11())

Hope that helps.

转载请标明出处:Deploy multiple web services, i.e. multiple wsdl files, in python

文章来源: Deploy multiple web services, i.e. multiple wsdl files, in python