Question

我有这个代码，获取用户名和密码。但问题是即使输入字段保持为空，它仍然可以运行。我希望如果输入字段为空，则用户必须再次输入用户名和密码。

from Tkinter import *
def onclick():
pass

import tkMessageBox

root = Tk() 

root.title("Pantai Hospital")
root.geometry("200x200") 
L1 = Label(root, text='Welcome to Pantai Hospital!')
L1.pack()

def messageWindow1():
   win = Toplevel()
   win.geometry("300x300") 
   frame1 = Frame(win)
   frame1.pack()

   frame2 = Frame (win)
   frame2.pack()

   frame3 = Frame (win)
   frame3.pack() 

   L2 = "Login"
   Label(frame1, text = L2).pack()

   L3 = Label(frame2, text = "Username:")
   L3.pack( side = LEFT, padx = 5, pady = 10)
   username = StringVar()
   E1 = Entry(frame2, textvariable = username, width = 40)
   E1.pack( side = LEFT)

   L4 = Label(frame3, text = "Password:")
   L4.pack( side = LEFT, padx = 5, pady = 10)
   password = StringVar()
   E2 = Entry(frame3, textvariable =  password, show = "*", width = 40)
   E2.pack( side = LEFT ) 

validuser = ["yeojin" , "1234"]
v = 0
 def loginCallBack() :
      counter = 0
    if username.get() and password.get() in validuser :
       tkMessageBox.showinfo( "Welcome!" , "You are successfully logged on.")
    else :
        tkMessageBox.showinfo ("Login error", "please try again.") 
        while (counter < 3):
            if username.get() and password.get() in validuser:
                tkMessageBox.showinfo( "Welcome!" , "You are successfully logged on." )
            else :
                counter = counter + 1

B3 = Button ( win, text = "Login", command = loginCallBack)
B3.pack()

有没有人有任何想法？

Answer 1

counter = 0

def loginCallBack() :
    global counter
    if username.get() == "" or password.get() == "":
        tkMessageBox.showinfo("Login error", "Username and password fields can not be empty.")
    elif username.get() == validuser[0] and password.get() == validuser[1]:
       tkMessageBox.showinfo( "Welcome!" , "You are successfully logged on.")
    else :
        if counter < 3:
            tkMessageBox.showinfo ("Login error", "Please try again.") 
            counter += 1
        else:
            #do whatever you want after 3 failed attempts
            root.destroy()

你的问题在这一行：

if username.get() and password.get() in validuser:

Python解释器将该行视为

if (username.get()) and (password.get() in validuser)

因此，当您链接条件时，您需要单独检查它们。像

这样的东西

if username.get() in validuser and password.get() in validuser。

如果您这样做，用户可以通过输入“1234”或“yeojin”字段登录，因此您应该专门检查这样的值。

if username.get() == validuser[0] and password.get() == validuser[1]

此外，要捕获空条目字段，您可以向loginCallBack()添加另一个条件。如果其中一个输入字段为空，则可以弹出消息而不执行任何其他操作。

除此之外，使用while循环不是一个好主意，因为那里已经存在无限循环（主循环）。
您可以在回调中创建外的全局计数器，并且在每次失败的尝试增量时，该值和每次单击都会检查该值是否小于3。

Python GUI中的输入字段不能为空

1 个答案: