在 Python 中获取正确的 Checkbutton 状态时遇到问题

Question

作为 Python 中的一个完全菜鸟，我在过去几天开始尝试使用语法来尝试创建某种想到的程序。我最初的想法是创建一个地方来创建您自己的密码，并让代码检查它是否正确（1 个符号、大写字母、数字），如果正确，它将被接受。所以我扩大了一点。

我的问题是我放了一个复选框并拼命尝试获取它的状态，这样每当它被选中时，就会显示密码，如果没有't，通行证会被解密为'*'。

from tkinter import *

root = Tk()
root.resizable(width=FALSE, height=FALSE)

#
# Function checks if password meets the criteria: No space, 1 symbol, 1 number, 1 capital letter.
# Makes changes once criteria is met.

def checkPass(event):
    global b
    b = entryPass.get()
    zDigit = sum(map(b.count, ("0", "1", "2", "3", "4", "5", "6", "7", "8", "9")))
    zSymbol = sum(map(b.count, ("!", "@", "#", "$", "%", "^", "&", "*", "(", ")", "-", "_", "=", "+", "[", "{", "]", "}", ";", ":", "'", ",", ">", ",", "<", "/", "|")))
    zCapital = sum(map(b.count, ("A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z")))
    zSpace = b.count(" ")
    if zCapital > 0:
        if zSymbol > 0:
            if zDigit > 0:
                if zSpace > 0:
                    print("You cannot include a space in your password!")
                else:
                    print("Your password is suitable and has been created!")
                    buttonCreate.grid_forget()
                    entryPass.grid_forget()
                    c = len(b)
                    global d
                    d = "*" * c
                    global firmLabel
                    firmLabel = Label(root, text=d)
                    firmLabel.grid(row=2, column=1)
                    baseLabel['text'] = 'SUCCESS'
                    baseLabel.grid(columnspan=1)
                    password.grid(row=2, sticky=E)
                    firmLabel['text'] = d
                    showPassBtn.grid(row=3, columnspan=2)
            else:
                print("You need at least one number in your password!")
        else:
            print("You need at least one symbol in your password!")
    else:
        print("You need at least one capital letter in your password!")

#
# Function that is called when the checkbox is clicked to show/hide password.

def showPass(btnState):
    if btnState == TRUE:
        firmLabel['text'] = d
    else:
        firmLabel['text'] = b

#
# The labels/buttons/inputs are set onto the main GUI (root).

# Checkbutton
btnState = IntVar()

showPassBtn = Checkbutton(root, text="Show password", variable=btnState)
showPassBtn.bind('<Button-1>', showPass)

# Rest of GUI Elements

baseLabel = Label(root, text="Type a password")
baseLabel.grid(row=0, columnspan=2)

password = Label(root, text="Password: ")

entryPass = Entry(root)
entryPass.grid(row=2, column=1)

buttonCreate = Button(root, text="Create")
buttonCreate.bind('<Button-1>', checkPass)
buttonCreate.grid(columnspan=2)


root.mainloop()

我尝试了很多方法，通过不同的资源，其中大部分都给我错误，我无法完成任务。我尝试通过多种方式获取按钮的状态，但我现在已经忘记了（其中一种是我尝试获取 btnState 的状态，但它告诉我该事件没有属性 "get" 或类似的东西） , 所以我真的很难找到解决办法。

对不起，如果代码很乱，正如我所说，我是一个完全的新手，并没有复习所有的语法以便我可以优化我的程序。我知道有一些东西需要优化，比如字符检查系统，但现在它可以正常工作了。

P.S。我设法通过使用以下方法让它工作：

def showPass(btnState):
    if state == 0:
        firmLabel['text'] = d
        global state
        state = 1
    else:
        firmLabel['text'] = b
        state = 0

state = 1

但这只是避免了获取按钮状态并使用它来做某件事的概念。

提前致谢！

Answer 1

您的主要错误是 showPass 函数（绑定到鼠标事件）没有将 CheckButton 值作为参数，而是将事件本身作为参数。有关详细信息，请参阅 here。您应该将其替换为：

    def showPass(event)

然后您可以将 btnState 视为一个全局变量（您可以在代码的上方使用 btnState=IntVar() 定义它），然后使用它来检索 CheckButton 值。如果 btnState.get() == 0，则表示未选中 CheckButton。

在 Python 中获取正确的 Checkbutton 状态时遇到问题

Having issue getting proper Checkbutton state in Python

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