如何自動通知用戶他們忘記選擇操作？

我正在創建一個登錄系統，但我有一個我不知道如何處理的錯誤，情況是這樣的：我希望每個用戶都能夠在使用加密和不使用加密之間進行選擇。例如，某人輸入了正確的登錄資訊，但該人忘記選擇訊息型別，并且當該人按下 Enter 按鈕時，他們會收到一個錯誤，提示他們忘記選擇訊息型別。我該如何實施？代碼如下：

from tkinter import messagebox
from tkinter import *


window = Tk()

window.title('Login')
window.geometry('320x200')
window.resizable(True, True)

name = StringVar()
password = StringVar()



def crypt():     

    r = (lis.get(lis.curselection()))
    c = (lis.get(lis.curselection()))

    string_name = name.get()
    string_password = password.get()

    #r = (lis.get(lis.curselection()))
    #c = (lis.get(lis.curselection()))



    if string_name == 'John':
        if string_password == '6789':
            if r == 'Use encrypted':
                window.after(1000, lambda: window.destroy())
                
                print('Hello.')


    if string_name == 'John':
        if string_password == '6789':
            if r == 'Use decrypted':
                window.after(1000, lambda: window.destroy())

                print('Hello bro!')
            

            
    if string_name not in 'John':
        messagebox.showerror('Error', 'Error')
    elif string_password not in '6789':
        messagebox.showerror('Error', 'Error')

    elif r not in r:                                                        
        messagebox.showerror('Error', 'Oops, please crypt message')    #This Error

    elif string_name == 'John':
        messagebox.showerror('Error', 'Error')
    elif string_password == '6789':
        messagebox.showerror('Error', 'Error')
    
entry = Entry(window, textvariable=name, width=10)
entry.grid(column=1, pady=7, padx=4)

label = Label(window, text='Enter name: ')
label.grid(row=0, padx=1)

entry1 = Entry(window, textvariable=password, width=10, show='*')
entry1.grid(column=1, pady=7, padx=2)

label1 = Label(window, text='Enter password: ')
label1.grid(row=1, padx=1)

listbox = Listbox(window, selectmode=SINGLE, width=12, height=2)
listbox.grid(column=1, row=2, pady=7, padx=2)



r = ['Use encrypted']
c = ['Use decrypted']
lis = Listbox(window, selectmode=SINGLE, width=10, height=2)
lis.grid(column=1, row=2, pady=7, padx=2)
for i in r:
    lis.insert(END, i)  
for i in c:
    lis.insert(END, i)

label_crypto = Label(window, text='Encrypted/decrypted message: ', bg='black', fg='red')
label_crypto.grid(row=2)

button = Button(window, text='Enter', command=crypt)
button.grid(pady=30)


window.mainloop()

uj5u.com熱心網友回復：

正如我在評論中所建議的，改進變數的名稱將更好地讓您區分它們。

下面的代碼使用 try-catch 塊來檢測用戶尚未從串列框中選擇專案。如果您嘗試在尚未選擇的情況下從串列中獲取所選專案，Tkinter 將引發錯誤。

from tkinter import messagebox
from tkinter import *
import _tkinter


window = Tk()

window.title('Login')
window.geometry('320x200')
window.resizable(True, True)

name = StringVar()
password = StringVar()



def crypt():     

    try:
        user_encryption_selection = (encryption_listbox.get(encryption_listbox.curselection()))
    except _tkinter.TclError:
        messagebox.showerror('Error','User has not selected an encryption type')
        return
        

    string_name = name.get()
    string_password = password.get()

    if string_name == 'John':
        if string_password == '6789':
            if user_encryption_selection == 'Use decrypted':
                window.after(1000, lambda: window.destroy())

                print('Hello bro!')
        else:
            messagebox.showerror('Error', 'Error Password')
    else:
        messagebox.showerror('Error', 'Invalid Username')

    
entry = Entry(window, textvariable=name, width=10)
entry.grid(column=1, pady=7, padx=4)

label = Label(window, text='Enter name: ')
label.grid(row=0, padx=1)

entry1 = Entry(window, textvariable=password, width=10, show='*')
entry1.grid(column=1, pady=7, padx=2)

label1 = Label(window, text='Enter password: ')
label1.grid(row=1, padx=1)

listbox = Listbox(window, selectmode=SINGLE, width=12, height=2)
listbox.grid(column=1, row=2, pady=7, padx=2)



encryption_options = ['Use encrypted','Use decrypted']
encryption_listbox = Listbox(window, selectmode=SINGLE, width=10, height=2)
encryption_listbox.grid(column=1, row=2, pady=7, padx=2)
for i in encryption_options:
    encryption_listbox.insert(END, i)  


label_crypto = Label(window, text='Encrypted/decrypted message: ', bg='black', fg='red')
label_crypto.grid(row=2)

button = Button(window, text='Enter', command=crypt)
button.grid(pady=30)


window.mainloop()

我還洗掉了一些不必要的代碼。您的目標應該是只檢查一次用戶名/密碼/加密值，而不是在單獨的 if/elif/else 條件中多次檢查

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