[英]Python ServerSocket running from a separate Thread
我想在单独的线程中运行Python serversocket类。 为此,我修改了此代码以支持python线程:
原始文件
import SocketServer
class MyTCPHandler(SocketServer.BaseRequestHandler):
def handle(self):
self.data = self.request.recv(1024).strip()
print self.data
if __name__ == "__main__":
HOST, PORT = "192.168.65.1", 5001
server = SocketServer.TCPServer((HOST, PORT), MyTCPHandler)
server.serve_forever()
到我的文件:
import SocketServer
class MyTCPHandler(SocketServer.BaseRequestHandler):
def __init__(self):
self._running = True
def terminate(self):
self._running = False
def setup(self):
print('{}:{} connected'.format(*self.client_address))
def handle(self):
self.data = self.request.recv(1024).strip()
print self.data
def finish(self):
print('{}:{} disconnected'.format(*self.client_address))
#if __name__ == "__main__":
def run(self):
HOST, PORT = "192.168.65.1", 5001
server = SocketServer.TCPServer((HOST, PORT), MyTCPHandler)
server.serve_forever()
我使用以下命令在主文件中将其作为单独的线程运行
Thread(target=MyTCPHandler().run, args=()).start()
但是,当我运行主python文件时,出现此错误
----------------------------------------
Exception happened during processing of request from ('192.168.65.1', 37997)
----------------------------------------
Traceback (most recent call last):
File "/usr/lib/python2.7/SocketServer.py", line 295, in _handle_request_noblock
self.process_request(request, client_address)
File "/usr/lib/python2.7/SocketServer.py", line 321, in process_request
self.finish_request(request, client_address)
File "/usr/lib/python2.7/SocketServer.py", line 334, in finish_request
self.RequestHandlerClass(request, client_address, self)
TypeError: __init__() takes exactly 1 argument (4 given)
首先,我对支持线程的修改是否正确? 拜托,有人可以帮我一个忙吗?
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