如何在建立新连接时退出由socketio.start_background_task()创建的先前打开的函数

Question

每次当我从客户端刷新页面时，都会与flask服务器建立新的连接，并且它运行函数'backgroundFunction()'而不退出最近打开的函数，并且随着我一次又一次刷新页面，数字会增加。

from flask import Flask
from flask_socketio import SocketIO, send, emit
import socket
from time import sleep
import datetime
app = Flask(__name__)
app.config['SECRET_KEY'] = 'secret'
app.config['DEBUG'] = True
socketio = SocketIO(app , cors_allowed_origins="*" , async_mode = None , logger = False , engineio_logger = False)

def backgroundFunction():
while True:
data = "I am Data"
socketio.emit('data', data, broadcast=True)
socketio.sleep(2)

@socketio.on('connect')
def socketcon():
print('Client connected')
socketio.start_background_task(backgroundFunction)

if __name__ == ("__main__"):
socketio.run(app, port=5009)

Answer 1

查看Flask-SocketIO存储库中的示例代码，了解实现在事件第一次触发时启动的后台作业的一种可能方法。

代码在这里。以下是相关节选:

thread = None
thread_lock = Lock()
def background_thread():
"""Example of how to send server generated events to clients."""
count = 0
while True:
socketio.sleep(10)
count += 1
socketio.emit('my_response',
{'data': 'Server generated event', 'count': count})
@socketio.event
def connect():
global thread
with thread_lock:
if thread is None:
thread = socketio.start_background_task(background_thread)