请求标题
POST /testfunction.PHP http/1.1Accept: application/soap+xml,application/xml,text/xmlSOAPAction: "http://www.abc.com/testfunction#test"Host: soap.abc.comContent-Length: 461Connection: Keep-Alive
请求内容
<?xml version="1.0" enCoding="UTF-8"?><SOAP-ENV:Envelope xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/" xmlns:ns1="http://www.abc.com/testfunction"><SOAP-ENV:Body><ns1:test><productID>3</productID><productSubID>1</productSubID><version>1.0.1</version><serialNumber/><language>EN</language><dayslicensed>0</dayslicensed><daysleft>0</daysleft></ns1:test></SOAP-ENV:Body></SOAP-ENV:Envelope>
SOAP服务应该响应:
响应标题
http/1.1 200 OKServer: Nginx/0.7.67Date: Mon,02 May 2011 13:43:46 GMTContent-Type: text/xml; charset=utf-8Connection: keep-aliveX-Powered-By: PHP/5.3.3-1ubuntu9.3Content-Length: 304content-encoding: gzip
响应内容
<?xml version="1.0" enCoding="UTF-8"?><SOAP-ENV:Envelope xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/" xmlns:ns1="http://www.abc.com/testfunction"><SOAP-ENV:Body><ns1:testResponse><result>1000</result></ns1:testResponse></SOAP-ENV:Body></SOAP-ENV:Envelope>
一开始我以为我可以在web.py中创建一个Web服务,并在有人在http://www.abc.com/testfunction发出POST请求时返回相同的响应.
import web url = ( '/testfunction','testfunction',) class testfunction: def POST(self): web.header('Content-Type','text/xml; charset=utf-8') return """<?xml version="1.0" enCoding="UTF-8"?><SOAP-ENV:Envelope xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/" xmlns:ns1="http://www.abc.com/testfunction"><SOAP-ENV:Body><ns1:testResponse><result>1000</result></ns1:testResponse></SOAP-ENV:Body></SOAP-ENV:Envelope> """ app = web.application(url,globals()) if __name__== "__main__": app.run()
但它没有用.我想这可能与标题有关.然后我尝试使用SimpleXMLRPCServer.
from SimpleXMLRPCServer import SimpleXMLRPCServer,SimpleXMLRPCRequestHandlerimport xmlrpclibclass MyRequestHandler(SimpleXMLRPCRequestHandler): rpc_paths = ('/',) def do_POST(self): return SimpleXMLRPCRequestHandler.do_POST(self)def test(): return "hello,world"server = SimpleXMLRPCServer( ("0.0.0.0",80),requestHandler = MyRequestHandler,) server.register_function(test,'test')server.serve_forever()
但问题是我不知道如何在标题中处理SOAPAction,而客户端在这里没有使用测试函数.谁能帮我?非常感谢!!
更新:
终于做到了,使用以下代码:
from wsgiref.simple_server import WsgiServer,WsgiRequestHandlerdef echo(environ,start_response): status = "200 OK" headers = [("Content-type","text/xml")] start_response(status,headers) return ["""<?xml version="1.0" enCoding="UTF-8"?> <SOAP-ENV:Envelope xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/" xmlns:ns1="http://www.abc.com/testfunction"><SOAP-ENV:Body><ns1:testResponse><result>1000</result></ns1:testResponse></SOAP-ENV:Body></SOAP-ENV:Envelope>"""]httpd = WsgiServer(('0.0.0.0',WsgiRequestHandler)httpd.set_app(echo)httpd.serve_forever()
它应该与web.py代码一样,但web.py没有.从wireshark中捕获的包中,我在xml内容之前和之后发现了一些乱码(“3bc”和“0”),与编码有关?
解决方法 您可以使用 soaplib创建一个真正的SOAP服务,该服务实现您的接口并返回虚拟数据.这应该更容易维护,然后创建手写静态响应,并且代码不应该比基于web.py的示例更长.这是Hello World example.
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