tkinter：如何在同一处理程序中的小部件更改之间插入延迟？

Question

大约两年前，当我试图在 Tkinter 上模拟 LED 时，我问过一个类似的问题 canvas。解决方案是使用 canvas after() 方法而不是 sleep() 函数来引入小部件更新之间的延迟。

从那时起，我发现了tk_tools模块，它具有内置功能来创建LED，太棒了！但是我现在遇到了和以前一样的问题，就是：如何在每个LED的开启（变为绿色）之间有1秒的延迟？

当运行下面的代码实际发生的是 LED 显示为关闭状态（灰色），然后当我单击 'Start' 按钮时，有 4 秒的延迟所有 LED 同时亮起。

谢谢。约翰尼姆

# LED array simulation

from tkinter import *
import tk_tools as tkt
from time import *

# turn on LEDs (change to green) with a 1-second delay between each
def turn_on():
    for led in range(4):
        led_array[led].to_green()
        sleep(1)

# list to hold a 4-LED array
led_array = []

# GUI
root = Tk()

# create 4 LED widgets, store them in led_array[], display them
for i in range(4):
    led_array.append(tkt.Led(root, size=30))
    led_array[i].grid(row=0, column=i)

# create button to initiate LED turn-on sequence
start = Button(root, text='Start', padx=20, command=turn_on)
start.grid(row=1, columnspan=4)



root.mainloop()

Answer 1

我想每当回调函数完成执行时，所有 LED 都会同时发光。所以我决定创建一个 thread 并使用 after() 来停止时间（sleep() 停止主线程，使其持续 4 秒）

# LED array simulation

from tkinter import *
import tk_tools as tkt
from time import *
import threading
import sys

class Testing(Tk):
    def __init__(self):
        super().__init__()
        self.led_array = []
        for i in range(4):
            self.led_array.append(tkt.Led(self, size=30))
            self.led_array[i].grid(row=0, column=i+1)
        self.start = Button(self, text='Start', padx=20, command=self.zawardu)
        self.start.grid(row=1, columnspan=4)
    def zawardu(self):
        a=threading.Thread(target=self.turn_on)
        a.start()
    def turn_on(self):
        self.start.config(state=DISABLED) # to stop creating another thread unnec..
        for led in range(len(self.led_array)):
            self.led_array[led].to_green()
            self.after(1000) # stops time
        self.start.config(state=NORMAL) 
        sys.exit(0) # ensures after exec. thread is killed
# class format to use after() and threading to reset the button status
Testing().mainloop()

Answer 2

最好避免停止应用程序的睡眠并使用 after。要么将调用链接到后回调，以便每个调用下一个调用，要么在循环中为每个 led 调用它们，但在 1、2、3 和 4 秒延迟。

链接版本：

# LED array simulation

from tkinter import *
import tk_tools as tkt

# list to hold a 4-LED array
led_array = []

# GUI
root = Tk()

# create 4 LED widgets, store them in led_array[], display them
for i in range(4):
    led_array.append(tkt.Led(root, size=30))
    led_array[i].grid(row=0, column=i)

# Both answers are common to here.

def on_after( led_list ):
    led_list[0].to_green() # Set first item in the list.
    if len( led_list ) > 1:
        # Call on_after again with a shortened list.
        root.after( 1000, on_after, led_list[1:] )
    else:
        # Enable the start button
        start.config( state = NORMAL )

# turn on LEDs (change to green) with a 1-second delay between each
def turn_on():
    start.config( state = DISABLED ) # Disable the button
    root.after( 1000, on_after, led_array)

# create button to initiate LED turn-on sequence
start = Button(root, text='Start', padx=20, command=turn_on)
start.grid(row=1, columnspan=4)

root.mainloop()

循环版：

# Both answers are common to here.
def on_after( led ):
    led.to_green()

def enable():
    start.config( state = NORMAL )

# turn on LEDs (change to green) with a 1-second delay between each
def turn_on():
    start.config( state = DISABLED ) # Disable the button
    for ix, led in enumerate( led_array ):
        # Call on_after 4 times with 1 to 4 second delays
        root.after( 1000 * (1+ix), on_after, led )
    root.after( 1000*(ix+1), enable )

# create button to initiate LED turn-on sequence
start = Button(root, text='Start', padx=20, command=turn_on)
start.grid(row=1, columnspan=4)

root.mainloop()

我可能会使用链接版本，尽管在这种情况下循环版本可能更容易理解。

tkinter：如何在同一处理程序中的小部件更改之间插入延迟？

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