如何使用线程更新tkinter gui标签？

如何解决如何使用线程更新tkinter gui标签？

我是Python tkinter的新手。我已经为gui编写了以下代码。我想用从Rabbitmq收到的正文消息更新标签1。但是，一旦我的gui被填充，即使我在正文中收到不同的消息，我也面临问题，但是它无法更新。一旦我关闭了gui，那么它又有了新的价值。我希望我的gui tkinter窗口保持不变，并且在接收到正文中的新消息时应该刷新标签。

import tkinter
from PIL import ImageTk,Image as PILImage
import datetime as dt
from tkinter import *
import pika

connection = pika.BlockingConnection(
    pika.ConnectionParameters(host='localhost'))
channel = connection.channel()

channel.queue_declare(queue='hello')


def callback(ch,method,properties,body):
    global myval
    print(" [x] Received %r" % body)
   
    window=Tk()
    window.attributes('-fullscreen',True)
    window.bind("<F11>",lambda event: window.attributes("-fullscreen",not window.attributes("-fullscreen")))
    window.bind("<Escape>",False))
    top_left=Frame(window,width=200,height=200)
    top_middle=Frame(window,width=550,height=200)
    top_right=Frame(window,width=250,height=200)
    middle_left=Frame(window,height=300)
    middle_middle=Frame(window,width=300,height=300)
    middle_right=Frame(window,height=300)
    bottom_left=Frame(window,width=0,height=200)
    bottom_middle=Frame(window,height=200)
    bottom_right=Frame(window,height=200)
    top_left.grid(row=0,column=0)
    top_middle.grid(row=0,column=1)
    top_right.grid(row=0,column=2,sticky=E+W)
    middle_left.grid(row=1,column=0,padx=100,pady=100)
    middle_middle.grid(row=1,column=1) 
    middle_right.grid(row=1,column=2)
    bottom_left.grid(row=2,column=0)
    bottom_middle.grid(row=2,column=1)
    bottom_right.grid(row=2,column=2)
    dte=Label(top_left,text="Date: "f"{dt.datetime.now():%a,%d/ %m/ %Y}",fg="black",font=("Arial Bold ",12 ))
    dte.place(x=0,y=40)
    lbl=Label(top_middle,text="Welcome to Smartcards Division",fg='#3333ff',font=("Arial Bold Italic",24 ))
    lbl.place(x=0,y=30)
    logo_path="logo.jpg"
    logo = ImageTk.PhotoImage((PILImage.open(logo_path)).resize((280,100),PILImage.ANTIALIAS))
    logo_panel = Label(top_right,image = logo)
    logo_panel.place(x=10,y=30)
    string_clsname=str(body.decode())
    lblxt=StringVar()
    lbl1=Label(middle_left,textvariable=lblxt,fg='#ff6600',16))
    lblxt.set("Hello "+string_clsname+" Sir")
    lbl1.place(x=0,y=100)
    path = "NewPicture_Copy.jpg"
    image = ImageTk.PhotoImage((PILImage.open(path)).resize((250,250),PILImage.ANTIALIAS))
    panel = Label(middle_middle,image = image,borderwidth=5,relief="ridge")
    panel.pack()
    lbl2=Label(bottom_middle,text="\u00a9"+"2020-Smartcards Division",fg='black',font=("Helvetica",8))
    lbl2.place(x=0,y=0)
    window.title('Image Classification')
    window.mainloop()

channel.basic_consume(
    queue='hello',on_message_callback=callback,auto_ack=True)
print(' [*] Waiting for messages. To exit press CTRL+C')
channel.start_consuming()

解决方法

在基本级别上，您需要：

• 单独的执行线程，
  用于单独的任务（必须同时运行）。
• 线程彼此通信的一种方式；
  同时避免比赛条件
  （例如在一个线程中修改变量，
  而另一个线程正在读取它。
  在这里您可以例如使用互斥锁/锁，消息传递等。

import tkinter as tk

from collections import deque
from threading import Thread

from random import randint
from time import sleep

# Starting out (this is the main/gui thread).

root = tk.Tk()

label = tk.Label(root,text='Original text')
label.pack()

# Means of communication,between the gui & update threads:
messageQueue = deque()

# Create a thread,that will periodically emit text updates.
def emitText():  # The task to be called from the thread.
  while(True):  # Normally should check some condition here.
    messageQueue.append(f'Random number: {randint(0,100)}')
    sleep(1)  # Simulated delay (of 1 sec) between updates.
# Create a separate thread,for the emitText task:
thread = Thread(target=emitText)
# Cheap way to avoid blocking @ program exit: run as daemon:
thread.setDaemon(True)
thread.start()  # "thread" starts running independently.

# Moving on (this is still the main/gui thread).

# Periodically check for text updates,in the gui thread.
# Where 'gui thread' is the main thread,# that is running the gui event-loop. 
# Should only access the gui,in the gui thread/event-loop.
def consumeText():
  try: label['text'] = messageQueue.popleft()
  except IndexError: pass  # Ignore,if no text available.
  # Reschedule call to consumeText.
  root.after(ms=1000,func=consumeText)
consumeText()  # Start the consumeText 'loop'.

root.mainloop()  # Enter the gui event-loop.

另请参阅：

queue.Queue
“ collections.deque是
的替代实现 快速原子append（）和popleft（）的无界队列
不需要锁定的操作。“
collections.deque
threading.Thread