如何自动通知用户忘记选择操作？

Question

我正在创建一个登录系统，但我有一个错误，我不知道如何处理，情况是这样的:我希望每个用户都能在使用加密和不使用加密之间做出选择。例如，一个人输入了正确的登录信息，但他忘记了选择消息类型，当他按下Enter按钮时，他们会收到一个错误，说他们忘记选择消息类型。我如何实现这一点？代码如下：

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()

Answer 1

正如我在评论中所建议的，改进变量的名称将使您能够更好地区分它们。

下面的代码使用try-catch块来检测用户没有从列表框中选择项。Tkinter将抛出一个错误，如果您尝试从列表中获取所选项目，而没有选择一个项目。

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()

我还删除了一些不必要的代码。您的目标应该是只检查一次用户名/密码/加密值，而不是在单独的if/elif/else条件中多次检查