pythontkinter使用Raspberry Pi更新GUI

Question

最近我一直在用Raspberry pi开发Tkinter来构建一个家庭自动化的图形用户界面，我想设计一个验证模块，当传感器工作时，显示为“活动”，当传感器出现故障时，显示为“不活动”。 我可以在GUI中获得验证，但它不是动态的。每次我需要重新运行程序以获取传感器的更新状态

有没有一种方法可以在不重新运行整个程序的情况下更新活动和不活动状态？在

我从raspberry pi上的GPIO引脚获取输入，并在程序中读取它们

这是我目前为止所做的代码：

import RPi.GPIO as GPIO
import time
from tkinter import *
from tkinter import ttk
import tkinter.font


GPIO.setwarnings(False)


Sig1 = 7
GPIO.setmode(GPIO.BOARD)
GPIO.setup(Sig1, GPIO.IN)

out1 = 11# pin11
GPIO.setmode(GPIO.BOARD) # We are accessing GPIOs according to their physical location
GPIO.setup(out1, GPIO.OUT) # We have set our LED pin mode to output
GPIO.output(out1, GPIO.LOW)

gui = Tk()
gui.title("tkinter")
gui.config(background = "gray86")
gui.minsize(1050,620)

Font1 = tkinter.font.Font(family = 'Courier 10 Pitch', size = 20, weight = 'bold')
Font2 = tkinter.font.Font(family = 'Courier 10 Pitch', size = 18, weight = 'bold')
Font3 = tkinter.font.Font(family = 'Courier 10 Pitch', size = 9, weight = 'bold')

def func1_on():
    GPIO.output(out1, GPIO.HIGH) # led on
    Text1 = Label(gui,text=' ON ', font = Font2, bg = 'gray84', fg='green3', padx = 0)
    Text1.grid(row=8,column=1)

def func1_off():
    GPIO.output(out1, GPIO.LOW) # led off
    Text2 = Label(gui,text='OFF', font = Font2, bg = 'gray84', fg='red', padx = 0)
    Text2.grid(row=8,column=1)

label_2 = Label(gui,text='Sensor:', font = Font2, fg='gray40', bg = 'gray84', padx = 10, pady = 10)
label_2.grid(row=6,column=0)

if GPIO.input(Sig1) == True:
    Text3 = Label(gui,textvariable=' Active ',relief = "raised", font = Font2, bg = 'gray84', fg='green3', padx = 0)
    Text3.grid(row=6,column=1)
else:
    Text4 = Label(gui,textvariable=' Inactive ',relief = "raised", font = Font2, bg = 'gray84', fg='red', padx = 0)
    Text4.grid(row=6,column=1)


Button1 = Button(gui, text='Switch On', font = Font3, command = func1_on, bg='gray74', height = 1, width = 7)
Button1.grid(row=8,column=0)
Button2 = Button(gui, text='Switch Off', font = Font3, command = func1_off, bg='gray74', height = 1, width = 7)
Button2.grid(row=9,column=0)

gui.mainloop()

如果有人能帮助我，我将不胜感激。在

Answer 1

使用root.after(milliseconds, function_name)定期运行一些函数。在

此函数应检查传感器，更新标签中的文本，并在一段时间后使用root.after再次运行。在

顺便说一句：function_name应该没有()

---

最小的例子。它用当前时间更新标签

import tkinter as tk
import time

#  - functions  -

def check():
    # update text in existing labels
    label['text'] = time.strftime('%H:%M:%S')

    # run again after 1000ms (1s)
    root.after(1000, check) 

#  - main  -

root = tk.Tk()

label = tk.Label(root)
label.pack()

check() # run first time

root.mainloop()