Question

按下buttonGuess会运行numRandom函数，它也会运行remainingAttemps函数。问题是，如果用户按下buttonGues，则会重新分配Attemps值。

import tkinter
import random

window = tkinter.Tk()
window.geometry('600x500')

x = random.randint(1,10)
remainingTime = True
Attempts = 4

def countdown (time_left):
    global remainingTime
    if remainingTime == True:
        lblCrono.configure(text = str(time_left))
        if time_left > 0:
            time_left = time_left - 1
            window.after(1000, countdown, time_left)
        else:
            remainingTime = False
            lblCrono.configure(text = 0)
            return remainingTime, gameOver()
    else:
        return

def numRamdom():
    global Attempts
    numWritten = int(entryWriteNumber.get())
    if numWritten > x:
        lblClue.configure(text = 'Its a smaller number')
        return remainingAttempts(Attempts)
    if numWritten < x:
        lblClue.configure(text = 'Its a bigger number')
        return remainingAttempts(Attempts)
    if numWritten == x:
        lblClue.configure(text = 'Congratulations ;)')
        remainingTime = False
        return remainingTime, countdown(0)

def gameOver():
    if remainingTime == False and Attempts != 0:
        lblClue.configure(text = '¡Time\'s up!')
    else:
        lblClue.configure(text = 'No attempts')

def remainingAttempts(countAtempts):
    Attempts = countAtempts
    if Attempts == 0:
        return remainingTime, countdown(0), Attempts, gameOver()
    else:
        Attempts = Attempts - 1

entryWriteNumber = tkinter.Entry(window)
entryWriteNumber.grid(column = 0, row = 1, padx = 10, pady = 10)

lblNumber = tkinter.Label(window, text = 'Number', font = 'Comic 13 bold')
lblNumber.grid(column = 0, row = 0, padx = 10, pady = 10, sticky = tkinter.W )

buttonGuess = tkinter.Button(window, text = 'Guess', bg = 'light grey', padx = 20, command = numRamdom)
buttonGuess.grid(column = 0, row = 2, sticky = tkinter.W, padx = 10, pady = 10)

lblClue = tkinter.Label(window, text = 'Clue', font = 'Comic 13 bold')
lblClue.grid(column = 0, row = 3, padx = 10, pady = 10, sticky = tkinter.W )

lblCrono = tkinter.Label(window, text = '', bg = 'white', fg = 'red', font = 'Comic 20', padx = 50, pady = 5)
lblCrono.grid(column = 1, row = 5, sticky = tkinter.S, padx = 100, pady = 150)


countdown(30)


window.mainloop()

Answer 1

当您摆脱无所事事或不必要的一切时，管理起来会容易得多。您所有的returns都无所事事。有计数器时无需创建remainingTime变量。给小部件和功能提供一堆复杂和/或误导性的名称并没有帮助。您正在呼叫countdown(0)，后者先呼叫gameOver()，然后再呼叫gameOver()。

您从来没有在row=4上放任何东西，而是在row=5上放了计时器。显然，这与将其放到4上没有什么不同。您有非常重复的grid options，因此我在dict中将它们同质化，然后将dict用作**kwargs 。这样写参数-> func(arg1 = value1, arg2 = value2, ...)没有好处。没有理由保留对lblNumber或buttonGuess的引用。您绝不会以任何方式修改或进一步引用。如果您未指定column，则tkinter将假设您的意思是column=0。如果您未指定row，则tkinter会假设您的意思是比当前总行大1行，而与列无关。导入tkinter而不带别名只会给您带来更多输入。

以下是我根据我刚写的内容对游戏进行的编辑。

import tkinter as tk
import random

root = tk.Tk()
root.geometry('600x500')

x = random.randint(1,10)
remaining = 4

def countdown(time_left):
    global process
    chrono['text'] = str(time_left)
    if time_left:
        process = root.after(1000, countdown, time_left-1)
    else:
        gameOver()

def check():
    global remaining
    
    n = int(guess.get())
    if n == x:
        gameOver(True)
        return
    else:
        clue['text'] = f'Its a {"smaller" if n > x else "bigger"} number'
        
    remaining -= 1
    if not remaining:
        gameOver()

def gameOver(win=False):
    root.after_cancel(process)
    if not win:
        clue['text'] = '¡Time\'s up!' if remaining else 'No attempts remain'
    else:
        clue['text'] = 'Congratulations ;)'


grid = dict(padx=10, pady=10, sticky=tk.W)

tk.Label(root, text='Number', font='Comic 13 bold').grid(**grid)

guess = tk.Entry(root)
guess.grid(**grid)

tk.Button(root, text='Guess', bg='light grey', padx=20, command=check).grid(**grid)

clue = tk.Label(root, text='Clue', font='Comic 13 bold', width=20, anchor='w')
clue.grid(**grid)

chrono = tk.Label(root, text='', bg='white', fg='red', font='Comic 20', padx=50, pady=5)
chrono.grid(column=1, **grid)

countdown(30)

root.mainloop()

如何避免通过再次按下按钮来重置值

1 个答案: