需要帮助在两个套接字之间创建TCP中继
作者：互联网
我有以下情况：
SomeServer(S) <-> (C)MyApp(S) <-> (C)User
(S) represents a server socket
(C) represents a client socket
本质上,MyApp启动与SomeServer的通信(SomeServer(-)(C)MyApp),并且一旦某些身份验证例程成功,MyApp(S)开始等待(C)用户连接.用户连接后,MyApp会将数据从SomeServer中继到用户.这在两个方向上都发生.
我有SomeServer(S)<-> (C)MyApp正常运行,但无法获得MyApp(S)<-> (C)用户工作.我可以看到用户已连接到MyApp(S),但无法中继数据！
好的,我希望这很清楚；)现在让我展示我的MyApp代码.顺便说一句,SomeServer和User的实现与解决我的问题无关,因为两者都不能修改.
我已经注释了我的代码,指出了我在哪里遇到问题.哦,我还应该提到,如果有必要,我可以毫无问题地为整个其他“服务器部分”添加其他代码.这是一个POC,所以我的主要重点是使功能起作用而不是编写高效的代码.谢谢您的时间.
''' MyApp.py module '''
import asyncore, socket
import SSL
# Client Section
# Connects to SomeServer
class MyAppClient(asyncore.dispatcher):
def __init__(self, host, port):
asyncore.dispatcher.__init__(self)
self.create_socket(socket.AF_INET, socket.SOCK_STREAM)
self.connect((host, port))
connectionPhase = 1
def handle_read(self):
print "connectionPhase =", self.connectionPhase
# The following IF statements may not make sense
# as I have removed code irrelevant to this question
if self.connectionPhase < 3: # authentication phase
data = self.recv(1024)
print 'Received:', data
# Client/Server authentication is handled here
# Everything from this point on happens over
# an encrypted socket using SSL
# Start the RelayServer listening on localhost 8080
# self.socket is encrypted and is the socket communicating
# with SomeServer
rs = RelayServer(('localhost', 8080), self.socket)
print 'RelayServer started'
# connectionPhase = 3 when this IF loop is done
elif self.connectionPhase == 3: # receiving data for User
data = self.recv(1024)
print 'Received data - forward to User:', data
# Forward this data to User
# Don't understand why data is being read here
# when the RelayServer was instantiated above
# Server Section
# Connects to User
class RelayConnection(asyncore.dispatcher):
def __init__(self, client, sock):
asyncore.dispatcher.__init__(self)
self.client = client
print "connecting to %s..." % str(sock)
def handle_connect(self):
print "connected."
# Allow reading once the connection
# on the other side is open.
self.client.is_readable = True
# For some reason this never runs, i.e. data from SomeServer
# isn't read here, but instead in MyAppClient.handle_read()
# don't know how to make it arrive here instead as it should
# be relayed to User
def handle_read(self):
self.client.send(self.recv(1024))
class RelayClient(asyncore.dispatcher):
def __init__(self, server, client, sock):
asyncore.dispatcher.__init__(self, client)
self.is_readable = False
self.server = server
self.relay = RelayConnection(self, sock)
def handle_read(self):
self.relay.send(self.recv(1024))
def handle_close(self):
print "Closing relay..."
# If the client disconnects, close the
# relay connection as well.
self.relay.close()
self.close()
def readable(self):
return self.is_readable
class RelayServer(asyncore.dispatcher):
def __init__(self, bind_address, MyAppClient_sock):
asyncore.dispatcher.__init__(self)
self.create_socket(socket.AF_INET, socket.SOCK_STREAM)
self.bind(bind_address)
self.MyAppClient_sock = MyAppClient_sock
print self.MyAppClient_sock
self.listen(1)
def handle_accept(self):
conn, addr = self.accept()
RelayClient(self, conn, self.MyAppClient_sock)
if __name__ == "__main__":
# Connect to host
# First connection stage
connectionPhase = 1
c = MyAppClient('host', port) # SomeServer's host and port
asyncore.loop()
编辑：
@samplebias我用您的代码替换了完整的模块(未显示),并重新添加了身份验证等所需的所有细节.
在这一点上,我得到的结果与上面我自己的代码相同.我的意思是MyApp(或您代码中的Server)已连接到SomeServer并来回传递数据.到目前为止一切都很好.当用户(或客户端应用程序)连接到本地主机8080时,将运行以下代码：
if not self.listener:
self.listener = Listener(self.listener_addr, self)
但是,这没有运行
# if user is attached, send data
elif self.user:
print 'self.user'
self.user.send(data)
因此,服务器未将数据中继到用户.我在整个User类中添加了打印语句,以查看运行的内容和init是唯一的内容. handle_read()永远不会运行.
为什么是这样？
解决方法:
该代码很难遵循,我敢肯定有一些错误.对于
在handle_read()中的示例,您要将MyAppClient的原始套接字self.socket传递给
中继服务器.最后,MyAppClient和RelayConnection都在同一个套接字上工作.
而不是建议对我放在一起的原始代码进行错误修复
一个示例,它可以实现您的代码意图,并且更加简洁,易于理解.
我已经测试了它与IMAP服务器的通信情况,并且可以工作,但是省略了一些
为简洁起见(错误处理,在所有情况下都正确处理close()等).
>服务器启动与“ someserver”的连接.一旦连接
它启动监听器.
>侦听器侦听端口8080并仅接受1个连接,创建一个用户,
并将其传递给服务器.侦听器拒绝所有其他
用户处于活动状态时与客户端连接.
>用户将所有数据转发到服务器,反之亦然.评论
指示应在何处插入身份验证.
资源：
import asyncore
import socket
class User(asyncore.dispatcher_with_send):
def __init__(self, sock, server):
asyncore.dispatcher_with_send.__init__(self, sock)
self.server = server
def handle_read(self):
data = self.recv(4096)
# parse User auth protocol here, authenticate, set phase flag, etc.
# if authenticated, send data to server
if self.server:
self.server.send(data)
def handle_close(self):
if self.server:
self.server.close()
self.close()
class Listener(asyncore.dispatcher_with_send):
def __init__(self, listener_addr, server):
asyncore.dispatcher_with_send.__init__(self)
self.server = server
self.create_socket(socket.AF_INET, socket.SOCK_STREAM)
self.set_reuse_addr()
self.bind(listener_addr)
self.listen(1)
def handle_accept(self):
conn, addr = self.accept()
# this listener only accepts 1 client. while it is serving 1 client
# it will reject all other clients.
if not self.server.user:
self.server.user = User(conn, self.server)
else:
conn.close()
class Server(asyncore.dispatcher_with_send):
def __init__(self, server_addr, listener_addr):
asyncore.dispatcher_with_send.__init__(self)
self.server_addr = server_addr
self.listener_addr = listener_addr
self.listener = None
self.user = None
def start(self):
self.create_socket(socket.AF_INET, socket.SOCK_STREAM)
self.connect(self.server_addr)
def handle_error(self, *n):
self.close()
def handle_read(self):
data = self.recv(4096)
# parse SomeServer auth protocol here, set phase flag, etc.
if not self.listener:
self.listener = Listener(self.listener_addr, self)
# if user is attached, send data
elif self.user:
self.user.send(data)
def handle_close(self):
if self.user:
self.user.server = None
self.user.close()
self.user = None
if self.listener:
self.listener.close()
self.listener = None
self.close()
self.start()
if __name__ == '__main__':
app = Server(('someserver', 143), ('localhost', 8080))
app.start()
asyncore.loop()
标签：asyncore,sockets,asynchronous,python 来源： https://codeday.me/bug/20191102/1994425.html