我正在使用Python2.7。
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MeasAgErrEver对象时,我希望立即执行客户端中的下一行代码,而不必等待MeaGeErrEver的退出代码。有办法吗?
最佳答案
self.run()
改为呼叫
start()
。run()
在当前线程中执行run方法。start()
旋转另一个线程并调用它。self.start()
关于python - 调用另一个脚本后,立即使父脚本继续,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/36034804/