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.

来源：https://stackoverflow.com/questions/20275836/deploy-multiple-web-services-i-e-multiple-wsdl-files-in-python