使用 Python 更新标签 Tkinter答案

【问题标题】：Using Python to update labels Tkinter使用 Python 更新标签 Tkinter
【发布时间】：2018-05-30 12:47:37
【问题描述】：

我正在制作一个刽子手类型的游戏。到目前为止，我已经编写了运行良好的 CLI 版本，我只是将其移植到创建 GUI。

我卡住了:-(。程序不完整，还有很多工作要做，但我有两个问题。第一个是标签更新。

When a letter is chosen, it creates enough dashes for the letter, and places this in a label called 'letter'.

当用户输入一个字母时，它会替换破折号，但是它会在旧标签旁边添加一个新标签，相反，我想替换这个标签。我曾尝试使用 .set() 但这似乎不起作用。

我的第二个问题，这更像是一个逻辑错误（我认为），是我想跟踪输入的字母，以便我可以将其与新输入的字母进行比较并提醒用户。这很好用，但是当输入一个字母时，即使是第一次输入它也会警告用户。

代码如下：

import tkinter
from tkinter import *
from tkinter import messagebox
import random

guesses = 8

def play():
    print("play game")

    wordList = ["talking", "dollar","choice", "famous", "define", "features"]

    wordChoice = random.choice(wordList)

    print(wordChoice)
    wordLength = (len(wordChoice))
    print(wordLength)
    guessedLetters = []
    dashes = []

    def enterLetter():
        print("Guess")
        global guesses

        print(guessedLetters)

        while guesses != 0:
            guess = entry.get().lower()

            if len(guess) >1:
                messagebox.showinfo("Error","Sorry, only one letter at a time")
                entry.delete("0","end")
                return
            elif guess.isalpha() == False:
                messagebox.showinfo("Error","Letters only please")
                entry.delete("0","end")
                return
            elif guess in guessedLetters:
                messagebox.showinfo("Error","You have already used the letter")
                entry.delete("0","end")
                return

            guessedLetters.append(guess)
            print(guessedLetters)
            print(guesses)
            count = 0


            for i in range(wordLength):
                if wordChoice[i] == guess:
                    dashes[i] = guess
                    count = count +1
                    letter = Label(play, text = dashes, font = ("Arial",20)).grid(row = 2, column = i+1,padx = 10, pady =10)

            if count == 0:
                guesses -= 1

            if guesses == 0:
                print("You have ran out of guesses!")
                print("The word was:",wordChoice)

    ######  Play Game GUI


    play = Toplevel()
    play.title("Play Hangman!")

    label = Label(play, text="HANGMAN", font = ("Arial",16)).grid(row = 0)
    label = Label(play, text="Enter your guess >").grid(row = 3, column = 0)

    for i in range(wordLength):
        dashes.append("-")

    letter = Label(play, text = dashes, font = ("Arial",20)).grid(row = 2, column = i+1,padx = 10, pady =10)


    entry = Entry(play)
    entry.grid(row = 3, column = 1, columnspan = wordLength) 
    enterButton = Button(play, text = "Enter Guess", width = 15, command = enterLetter).grid(row = 3, column = (wordLength+2))
    label = Label(play, text = "Letter used: ").grid(row = 4, columnspan = 2)
    label = Label(play, text = "").grid(row= 4, columnspan = 6)



def scores():
    print("check scores")

def howToPlay():
    print("how to play")



####### Main Menu

root = Tk()
root.geometry("500x300")                                                                    
root.title("HANGMAN")
label = Label(root, text="HANGMAN", font = ("Arial",30)).grid(row = 0, columnspan = 3)
label = Label(root, text = "Option 1 :", font = ("Arial",12)).grid(row = 1, column = 1)    
playButton = Button(root, text = "Play Game", width = 15, command = play).grid(row = 1, column = 2) 
label = Label(root, text = "Option 2 :", font = ("Arial",12)).grid(row = 2, column = 1)
instructionsButton = Button(root, text = "How to play", width = 15, command = howToPlay).grid(row = 2, column = 2) 
label = Label(root, text = "Option 3 :", font = ("Arial",12)).grid(row = 3, column = 1) 
scoresButton = Button(root, text = "View Scores", width = 15, command = scores).grid(row = 3, column = 2) 
label = Label(root, text = "Option 4 :", font = ("Arial",12)).grid(row = 4, column = 1)
exitButton = Button(root, text = "Exit", width = 15, command = exit).grid(row = 4, column = 2)


root.mainloop()

【问题讨论】：

标签： python tkinter label

【解决方案1】：

您需要配置标签，而不是重新创建它。
为什么在 enter_letter 中使用 while 循环？它只是在单击按钮时运行，它需要是if guesses > 0:
当输入正确的单词时您的程序没有终止；我添加了这个。

代码：

import tkinter
from tkinter import *
from tkinter import messagebox
import random

guesses = 8

letter = None

def play():
    global letter

    print("play game")

    wordList = ["talking", "dollar","choice", "famous", "define", "features"]

    wordChoice = random.choice(wordList)

    print(wordChoice)
    wordLength = (len(wordChoice))
    print(wordLength)
    guessedLetters = []
    dashes = []


    play = Toplevel()
    play.title("Play Hangman!")

    label = Label(play, text="HANGMAN", font = ("Arial",16)).grid(row = 0)
    label = Label(play, text="Enter your guess >").grid(row = 3, column = 0)

    for i in range(wordLength):
        dashes.append("-")

    letter = Label(play, text = dashes, font = ("Arial",20))
    letter.grid(row = 2, column = i+1,padx = 10, pady =10)
    print(letter)

    def enterLetter():
        print("Guess")
        global guesses, letter

        print(guessedLetters)

        if guesses != 0:
            guess = entry.get().lower()

            if len(guess) >1:
                messagebox.showinfo("Error","Sorry, only one letter at a time")
                return
            elif guess.isalpha() == False:
                messagebox.showinfo("Error","Letters only please")
                return
            elif guess in guessedLetters:
                messagebox.showinfo("Error","You have already used the letter")
                return
            entry.delete("0","end")
            guessedLetters.append(guess)
            #print(guessedLetters)
            #print(guesses)
            print(dashes)
            count = 0


            for i in range(wordLength):
                if wordChoice[i] == guess:
                    dashes[i] = guess
                    count += 1
                    letter.configure(text = dashes)


            if count == 0:
                guesses -= 1

            if "".join(dashes) == wordChoice:
                print("succsess!")
                play.destroy()
                return

            if guesses == 0:
                print("You have ran out of guesses!")
                print("The word was:",wordChoice)

    ######  Play Game GUI


    entry = Entry(play)
    entry.grid(row = 3, column = 1, columnspan = wordLength) 
    enterButton = Button(play, text = "Enter Guess", width = 15, command = enterLetter).grid(row = 3, column = (wordLength+2))
    label = Label(play, text = "Letter used: ").grid(row = 4, columnspan = 2)
    label = Label(play, text = "").grid(row= 4, columnspan = 6)



def scores():
    print("check scores")

def howToPlay():
    print("how to play")



####### Main Menu

root = Tk()
root.geometry("500x300")                                                                    
root.title("HANGMAN")
label = Label(root, text="HANGMAN", font = ("Arial",30)).grid(row = 0, columnspan = 3)
label = Label(root, text = "Option 1 :", font = ("Arial",12)).grid(row = 1, column = 1)    
playButton = Button(root, text = "Play Game", width = 15, command = play).grid(row = 1, column = 2) 
label = Label(root, text = "Option 2 :", font = ("Arial",12)).grid(row = 2, column = 1)
instructionsButton = Button(root, text = "How to play", width = 15, command = howToPlay).grid(row = 2, column = 2) 
label = Label(root, text = "Option 3 :", font = ("Arial",12)).grid(row = 3, column = 1) 
scoresButton = Button(root, text = "View Scores", width = 15, command = scores).grid(row = 3, column = 2) 
label = Label(root, text = "Option 4 :", font = ("Arial",12)).grid(row = 4, column = 1)
exitButton = Button(root, text = "Exit", width = 15, command = exit).grid(row = 4, column = 2)


root.mainloop()

希望对你有帮助。

【讨论】：

感谢您提出使用 .join 的想法。我自己不打算使用它，但我想我现在就试试。 :-)

【解决方案2】：

您无法更新标签的原因是您没有将其存储在任何变量中。 Label 对象和所有其他小部件的 grid、pack 和 place 函数返回 None，因此当您调用时：

letter = Label(play, text = dashes, font = ("Arial",20)).grid(row = 2, column = i+1,padx = 10, pady =10)

您的标签无法通过可变字母访问。要解决此问题，您应该像这样拆分它：

letter = Label(play, text = dashes, font = ("Arial",20))
letter.grid(row = 2, column = i+1,padx = 10, pady =10)

要更新该标签的文本，您可以在其上调用.configure(text = 'new text')。

# letter = Label(play, text = dashes, font = ("Arial",20)).grid(row = 2, column = i+1,padx = 10, pady =10)    # 

letter.configure(text = dashes)

至于您的第二个问题，我认为您混淆了函数enterLetter 中的while 循环和if 语句。每次点击调用一次，如果玩家猜不到，您只需检查一次。

if guesses != 0:
    ...

elif guesses == 0:
    ....

【讨论】：