如何自动通知用户他们忘记选择操作

Question

我正在创建一个登录系统，但我遇到了一个不知道如何处理的错误，情况是：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，但忘记选择消息类型，当该人按下Enter按钮时，他们会收到一个错误，即忘记选择消息的类型。我该如何实现？以下代码：

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

正如我的评论中所建议的，改进变量的名称将更好地区分它们。

下面的代码使用try-catch块来检测用户是否没有从列表框中选择项目。如果您试图从列表中获取所选项目，而尚未选择该项目，Tkinter将抛出错误。

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。如果/elif/else条件，您应该只检查用户名/密码/加密值一次，而不是多次