在python中如何交替切换连接和断开mqtt

Question

因此，我想知道是否可以在连接到代理之间切换，获取一些消息，断开连接，执行一些其他代码，然后再连接到代理，然后从那里重复所有的过程。

我想做一个简单的闩锁。但不起作用。以下是代码：

时间锁存器= 0

import paho.mqtt.client as mqtt
import time

def on_connect(client, userdata, flags, rc):  # The callback for when the client connects to the broker
    print("Connected with result code {0}".format(str(rc)))  # Print result of connection attempt
    client.subscribe("trilaterasi/ESP1")
    client.subscribe("trilaterasi/ESP2")
    client.subscribe("trilaterasi/ESP3")
    
def on_message(client, userdata, msg):
    if msg.topic == 'trilaterasi/ESP1':
        global ESP1
        ESP1 = float(msg.payload)
        print("Msg rcvd" + msg.topic + " " + str(ESP1))
    elif msg.topic == 'trilaterasi/ESP2':
        global ESP2
        ESP2 = float(msg.payload)
        print("Msg rcvd" + msg.topic + " " + str(ESP2))
    elif msg.topic == 'trilaterasi/ESP3':
        global ESP3
        ESP3 = float(msg.payload)
        print("Msg rcvd" + msg.topic + " " + str(ESP3))
        
broker = "ipaddress"
latch = 0
while latch == 0:
    client = mqtt.Client("Python")
    client.loop_start()
    client.on_connect = on_connect
    client.on_message = on_message
    client.connect(broker)
    time.sleep(5)
    client.disconnect()
    client.loop_stop()
    latch = 1

时间锁存器= 1

while latch == 1:
    #This is the coordinates of ESPs that are mounted to the wall
    ESP1x = 300
    ESP1y = 0
    ESP2x = 0
    ESP2y = 130
    ESP3x = 150
    ESP3y = 260
    #This is the Trilateration Formula
    a = 2*(-ESP1x + ESP2x)
    b = 2*(-ESP1y + ESP2y)
    c = (ESP1**2 - ESP2**2 - ESP1x**2 + ESP2x**2 - ESP1y**2 + ESP2y**2)
    d = 2*(-ESP2x + ESP3x)
    e = 2*(-ESP2y + ESP3y)
    f = (ESP2**2 - ESP3**2 - ESP2x**2 + ESP3x**2 - ESP2y**2 + ESP3y**2)
    #This is the estimated coordinate of the BLE device
    x = ((c*e-f*b)/(e*a-b*d))
    y = ((c*d-a*f)/(b*d-a*e))
    print(x)
    print(y)
    latch = 0

它给出了结果，但它只运行了一次代码。我对python和mqtt非常陌生，所以我不知道自己做错了什么。我想知道是否因为while循环并尝试将其更改为if，但它仍然只运行了一次代码。我觉得我在这里错过了什么，所以任何建议都是非常感谢的！

Answer 1

如果我正确地理解了您的问题，那么您就有两个while循环：

latch = 0
while latch == 0:
   # Do stuff
   latch = 1
while latch == 1:
   # Do other stuff
   latch = 0

为什么每个循环只运行一次？如果您希望代码持续运行，那么需要一个外部循环。

latch = 0
while True:
   while latch == 0:
      # Do stuff
      latch = 1
   while latch == 1:
      # Do other stuff
      latch = 0

在这种情况下，实际上不需要latch，因为latch只(而且总是)在每个段的末尾被更改(因此while只运行一次)。您可以简化为：

while True:
   # Do stuff from first while loop (connect to MQTT broker etc)
   # Do stuff from second while loop (formula etc)

但是，需要重新考虑应用程序的结构。连接到代理涉及一些开销(建立连接、MQTT CONNECT握手，在您的情况下是SUBSCRIBE数据包等)，因此通常最好保持连接处于打开状态(而不是经常断开连接)。这有一个额外的好处:丢失更少的消息(另一种解决此问题的方法是通过setting the QOS)。也许可以这样做(粗略的伪代码)：

def recalculate():
    #This is the coordinates of ESPs that are mounted to the wall
    ESP1x = 300
    # Do calculation required when values change...

def on_message(client, userdata, msg):
    if msg.topic == 'trilaterasi/ESP1':
        global ESP1
        ESP1 = float(msg.payload)
        print("Msg rcvd" + msg.topic + " " + str(ESP1))
        recalculate()
    # Handle other messages 

client = mqtt.Client("Python")
client.on_connect = on_connect
client.on_message = on_message
client.connect(broker)

# Blocking call that processes network traffic, dispatches callbacks and
# handles reconnecting.
client.loop_forever()