带有mod_wsgi的soaplib,没有Django,Cherypy或其他框架

| 我在网上检查了soaplib的python并得到了示例
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\"
这个例子很好用。但是我想用mod_wsgi在Apache中运行它。我检查了网,所有都带有django,cherrypy或pylone。是否可以在没有任何Python网络框架的情况下运行此示例?在Apache的mod_wsgi下运行此示例需要遵循哪些步骤。我想在Unix中运行它。     
已邀请:
        与Wiki中的所有其他\“ Integration With \\”文档一样,除了
application = wsgi.Application(soap_application)
。     

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