soaplib与mod_wsgi没有django，cherypy或其他框架

Question

我在网上检查了soaplib for python，我得到了例子

import soaplib
from soaplib.core.service import rpc, DefinitionBase
from soaplib.core.model.primitive import String, Integer
from soaplib.core.server import wsgi
from soaplib.core.model.clazz import Array


class HelloWorldService(DefinitionBase):
    @soap(String,Integer,_returns=Array(String))
    def say_hello(self,name,times):
        results = []
        for i in range(0,times):
            results.append('Hello, %s'%name)
        return results

if __name__=='__main__':
    try:
        from wsgiref.simple_server import make_server
        soap_application = soaplib.core.Application([HelloWorldService], 'tns')
        wsgi_application = wsgi.Application(soap_application)
        server = make_server('localhost', 7789, wsgi_application)
        server.serve_forever()
    except ImportError:
        print "Error: example server code requires Python >= 2.5"

这个例子工作正常。但我想用apache中的mod_wsgi来运行它。我检查了网，所有都带有django，cherrypy或pylone。是否可以在没有任何python web框架的情况下运行此示例？在apache中的mod_wsgi下运行此示例需要遵循的步骤是什么。我想在unix中运行它。

Answer 1

与wiki中的所有其他“与...整合”文档一样，application = wsgi.Application(soap_application)除外。