Tkinter循环“Tkinter没有响应”

Posted 2021-05-01

我有一个来自与arduino通信的tkinter图形界面的代码，碰巧我得到时间打开和关闭的时间，所以我将每个当前时间与无限循环中的输入进行比较，该循环仅用于第二次在这里输入代码，要求关闭LED结束，它发生在进入循环时tkinter没有响应。

from tkinter import *
import serial
from time import strftime

conexao = serial.Serial('COM3', 9600, timeout=0.5)
cont=0
janela = Tk()
def ligar():
     global cont
     cont=cont+1
     if cont==1:
         valor=bytes(('1'),'utf-8')
         conexao.write(valor)
         ligar['text']='Desligar led'
     elif cont==2:
         valor=bytes(('2'),'utf-8')
         conexao.write(valor)
         ligar['text']='Ligar led'
         cont=0


def agendar():
     comeco= start.get()
     final=  limit.get()
     while 1>0:
          if comeco == strftime('%H:%M:%S'):
               valor=bytes(('1'),'utf-8')
               conexao.write(valor)
          elif final == strftime('%H:%M:%S'):
               valor=bytes(('2'),'utf-8')
               conexao.write(valor)
               break





ligar=Button(janela,command=ligar,text='ligar',width="60")
ligar.grid(row=1)
texto=Label(text='Defina o começo:').grid(row=2)
start=Entry(janela)
start.grid(row=3)
texto2=Label(text='Defina o fim:').grid(row=4)
limit=Entry(janela)
limit.grid(row=5)
salvar=Button(janela,width="30",text='Salvar Horarios',command=agendar)
salvar.grid(row=6)

答案

您可以使用线程来代替使用after方法，这允许您的GUI与while循环同时运行。

import threading
from tkinter import *
import serial
from time import strftime

conexao = serial.Serial('COM3', 9600, timeout=0.5)
cont=0
janela = Tk()
def ligar():
     global cont
     cont=cont+1
     if cont==1:
         valor=bytes(('1'),'utf-8')
         conexao.write(valor)
         ligar['text']='Desligar led'
     elif cont==2:
         valor=bytes(('2'),'utf-8')
         conexao.write(valor)
         ligar['text']='Ligar led'
         cont=0

def agendar():
     global start,limit
     comeco= start.get()
     final=  limit.get()
     while 1>0:
          if comeco == strftime('%H:%M:%S'):
               valor=bytes(('1'),'utf-8')
               conexao.write(valor)
          elif final == strftime('%H:%M:%S'):
               valor=bytes(('2'),'utf-8')
               conexao.write(valor)
               break

global start,limit
ligar=Button(janela,command=ligar,text='ligar',width="60")
ligar.grid(row=1)
texto=Label(text='Defina o começo:').grid(row=2)
start=Entry(janela)
start.grid(row=3)
texto2=Label(text='Defina o fim:').grid(row=4)
limit=Entry(janela)
limit.grid(row=5)
salvar=Button(janela,width="30",text='Salvar Horarios',command=lambda: threading.Thread(target=agendar).start())
salvar.grid(row=6)

这是一个简短的细分：

1. 按下按钮执行lambda函数（使用lambda函数很有用，因为它允许你避免使用包装函数）
2. lambda函数为agendar函数创建一个单独的线程
3. 线程开始

注意：我定义了limit并作为全局变量启动，但我不确定你是否需要它。 另外，我还没有测试过，但我相信它确实有效。

另一答案

您应该使用after定期运行代码。

我无法测试，但我会这样做

from tkinter import *
import serial
from time import strftime

# -- 
def ligar():
     global cont

     cont = not cont

     if cont:
         conexao.write( b'1' )
         ligar['text'] = 'Desligar led'
     else:
         conexao.write( b'2' )
         ligar['text'] = 'Ligar led'


def agendar():
    comeco = start.get()
    final  = limit.get()

    current = strftime('%H:%M:%S')

    if comeco == current:
        conexao.write( b'1' )
        # run again after 1000ms (1s)
        janela.after(1000, agendar)        
    elif final == current:
        conexao.write( b'2' )

conexao = serial.Serial('COM3', 9600, timeout=0.5)
cont = False

janela = Tk()

ligar = Button(janela, command=ligar, text='ligar', width="60")
ligar.grid(row=1)

texto = Label(janela, text='Defina o começo:')
texto.grid(row=2)

start = Entry(janela)
start.grid(row=3)

texto2 = Label(janela, text='Defina o fim:')
texto2.grid(row=4)

limit = Entry(janela)
limit.grid(row=5)

salvar = Button(janela, width="30", text='Salvar Horarios', command=agendar)
salvar.grid(row=6)

janela.mainloop()