用python实现有界面的用户登录功能,登陆后界面销毁
时间: 2023-06-18 22:05:56 浏览: 73
可以使用Python的GUI库Tkinter来实现带界面的用户登录功能。具体步骤如下:
1. 导入Tkinter和tkMessageBox模块
```python
from tkinter import *
import tkinter.messagebox as messagebox
```
2. 创建登录窗口,包括账号和密码输入框、登录和退出按钮
```python
root = Tk()
root.title('用户登录')
root.geometry('300x200')
label1 = Label(root, text='账号:')
label1.grid(row=0, column=0)
entry1 = Entry(root)
entry1.grid(row=0, column=1)
label2 = Label(root, text='密码:')
label2.grid(row=1, column=0)
entry2 = Entry(root, show='*')
entry2.grid(row=1, column=1)
button1 = Button(root, text='登录', command=login)
button1.grid(row=2, column=0, pady=10)
button2 = Button(root, text='退出', command=root.quit)
button2.grid(row=2, column=1, pady=10)
```
3. 编写登录函数,验证账号密码是否正确,如果正确则销毁登录窗口,否则弹出错误提示框
```python
def login():
if entry1.get() == 'admin' and entry2.get() == '123456':
messagebox.showinfo('提示', '登录成功')
root.destroy()
else:
messagebox.showerror('错误', '账号或密码错误')
```
4. 运行程序
```python
root.mainloop()
```
完整的代码如下:
```python
from tkinter import *
import tkinter.messagebox as messagebox
def login():
if entry1.get() == 'admin' and entry2.get() == '123456':
messagebox.showinfo('提示', '登录成功')
root.destroy()
else:
messagebox.showerror('错误', '账号或密码错误')
root = Tk()
root.title('用户登录')
root.geometry('300x200')
label1 = Label(root, text='账号:')
label1.grid(row=0, column=0)
entry1 = Entry(root)
entry1.grid(row=0, column=1)
label2 = Label(root, text='密码:')
label2.grid(row=1, column=0)
entry2 = Entry(root, show='*')
entry2.grid(row=1, column=1)
button1 = Button(root, text='登录', command=login)
button1.grid(row=2, column=0, pady=10)
button2 = Button(root, text='退出', command=root.quit)
button2.grid(row=2, column=1, pady=10)
root.mainloop()
```