Python - tkinter畫布內的倒數計時器

Question

你好，我想在子程序中創建一個倒數計時器，然後顯示在畫布上。我不完全確定從哪裏開始我已經對它做了一些研究，並且能夠用time.sleep（x）函數創建一個函數，但是該方法凍結了整個程序，這不是我之後。我還在這裏查詢了關於計時器的其他問題，並嘗試將它們合併到我的程序中，但我還沒有取得任何成功。

TLDR;我想創建一個倒數計時器，從60秒倒計時並顯示在畫布上，然後在計時器達到0時執行某些操作。

是否有人能夠將我指向正確的方向？

在此先感謝。

編輯：隨着提供的建議，我試圖把他們沒有多少運氣的程序。

不知道這段代碼是否有重大錯誤，或者它只是一個簡單的錯誤。 我運行它時得到的錯誤在代碼下面。

這是我想在計時器的部分代碼：

def main(): #First thing that loads when the program is executed. 
    global window 
    global tkinter 
    global canvas 
    global cdtimer 
    window = Tk() 
    cdtimer = 60 
    window.title("JailBreak Bob") 

    canvas = Canvas(width = 960, height = 540, bg = "white") 
    photo = PhotoImage(file="main.gif") 
    canvas.bind("<Button-1>", buttonclick_mainscreen) 
    canvas.pack(expand = YES, fill = BOTH) 
    canvas.create_image(1, 1, image = photo, anchor = NW) 
    window.mainloop() 

def buttonclick_mainscreen(event): 
    pressed = "" 

    if event.x >18 and event.x <365 and event.y > 359 and event.y < 417 : pressed = 1 
    if event.x >18 and event.x <365 and event.y > 421 and event.y < 473 : pressed = 2 
    if event.x >18 and event.x <365 and event.y > 477 and event.y < 517 : pressed = 3 

    if pressed == 1 : 
    gamescreen() 
    if pressed == 2 : 
    helpscreen() 
    if pressed == 3 : 
    window.destroy() 

def gamescreen(): 
    photo = PhotoImage(file="gamescreen.gif") 
    canvas.bind("<Button-1>", buttonclick_gamescreen) 
    canvas.pack(expand = YES, fill = BOTH) 
    canvas.create_image(1, 1, image = photo, anchor = NW) 
    game1 = PhotoImage(file="1.gif") 
    canvas.create_image(30, 65, image = game1, anchor = NW) 
    e1 = Entry(canvas, width = 11) 
    e2 = Entry(canvas, width = 11) 
    canvas.create_window(390, 501, window=e1, anchor = NW) 
    canvas.create_window(551, 501, window=e2, anchor = NW) 
    canvas.after(1, gamescreen) 
    window.mainloop() 

def cdtimer(): 
    canvas.delete(ALL) 
    global cdtimer 
    cdtimer -= 1 
    canvas.create_text(510, 6, text=cdtimer, font="Ubuntu 29 bold", anchor = NW) 
    if cdtimer == 0: 
    scorescreen() 
    else: 
    canvas.after(1000, gamescreen) 

main() 

錯誤味精：

Exception in Tkinter callback 
Traceback (most recent call last): 
    File "/usr/lib/python3.2/tkinter/__init__.py", line 1402, in __call__ 
    return self.func(*args) 
    File "/usr/lib/python3.2/tkinter/__init__.py", line 490, in callit 
    func(*args) 
    File "/home/ppppwn3d/workspace/Python/JailBreakBob/JailBreakBob.py", line 50, in  gamescreen 
    e1 = Entry(canvas, width = 11) 
    File "/usr/lib/python3.2/tkinter/__init__.py", line 2372, in __init__ 
    Widget.__init__(self, master, 'entry', cnf, kw) 
    File "/usr/lib/python3.2/tkinter/__init__.py", line 1952, in __init__ 
    cnf = _cnfmerge((cnf, kw)) 
    File "/usr/lib/python3.2/tkinter/__init__.py", line 71, in _cnfmerge 
    if isinstance(cnfs, dict): 
RuntimeError: maximum recursion depth exceeded while calling a Python object 

Answer 1

這是奧克利答案的擴展。它演示瞭如何顯示在畫布上的時間，以及揭開序幕，整個事情：

from tkinter import * 
root = Tk() 
canvas = Canvas(root) 
canvas.pack() 
time = 60 
def tick(): 
    # You have to clear the canvas each time the clock updates 
    # (otherwise it writes on top of the old time). Since the 
    # time is the only thing in the canvas, delete(ALL) works 
    # perfectly (if it wasn't however, you can delete the id 
    # that goes with the clock). 
    canvas.delete(ALL) 
    # I have to declare time as a global because I'm not using 
    # a class (otherwise, I could do something like self.time -= 1) 
    global time 
    time -= 1 
    # You can place the time wherever in the canvas 
    # (I chose 10,10 for the example) 
    canvas.create_text(10, 10, text=time) 
    if time == 0: 
     do_something() 
    else: 
     canvas.after(1000, tick) 
canvas.after(1, tick) 
root.mainloop() 

腳本計算從60秒（顯示爲它去剩餘時間），當它擊中0下來，調用do_something。

Answer 2

您要使用的after方法。邏輯是這樣的：

def update_clock(self): 
    self.counter -= 1 
    if self.counter == 0 : 
     do_something() 
    else: 
     self.after(1000, self.update_clock) 

以上將從計數器中減去一。如果計數器爲零，它會做一些特殊的事情。否則，它會安排自己在一秒鐘內再次運行。

Answer 3

你可能想嘗試線程？

import thread 
import time 

def myFunct(): 
    sec = 60 
    n = 1 
    for i in range(sec/n): 
     updateClock() 
     time.sleep(n) 
    finalFunct() 

thread.start_new_thread(myFunct,()) 

只要改變sec初始金額（單位：秒），並n到您希望它更新（以秒爲單位）的時間間隔。