带有mod_wsgi的soaplib,没有Django,Cherypy或其他框架
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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。
没有找到相关结果
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捕暑句簿姓
。