我使用的是 Python 2.7。
class Client():
def __init__(self, host, server_port):
"""
This method is run when creating a new Client object
"""
self.host = 'localhost'
self.server_Port = 1337
# Set up the socket connection to the server
self.connection = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
self.receiver = None
self.myParser = MessageParser()
# TODO: Finish init process with necessary code
self.run()
def run(self):
self.connection.connect((self.host, self.server_Port))
self.receiver = MessageReceiver(self, self.connection) #On this line, a MessageReceiver object is instantiated.
self.take_input()
class MessageReceiver(Thread):
def __init__(self, client, connection):
super(MessageReceiver, self).__init__()
self.myClient = client
self.connection = connection
self.daemon = True
self.run()
def run(self):
self.myClient.receive_message(self.connection.recv(1024)) #This line blocks further progress in the code.
当 Client 对象中的 run-方法实例化 MessageReceiver 对象时,我希望立即执行 Client 中的下一行代码,而不是等待 MessageReceiver 的退出代码。有办法做到这一点吗?
最佳答案
self.run()
改为调用start()
。 run()
在当前线程中执行run方法。 start()
启动另一个线程并在那里调用它。
self.start()
关于python - 使父脚本在调用另一个脚本后立即继续,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36034804/