我如何通过 lambda 函数处理异常？ [tkinter]

Question

大学生。我这里有一些代码，但我似乎无法从 lambda 函数中检索 .get() 用户条目。结果，我不能做任何异常处理。我唯一能解决的问题是检查以确保数据库中没有重复的用户名。我尝试将 .get() 变量返回给函数，但无济于事。这是我所指的代码片段：

def second_win():
    window = Tk()
    window.title("Please enter your personal details")
    window.geometry('500x500')
    Label(window, text = 'Create Username').grid(row = 0, column = 0)
    Label(window, text=" First Name").grid(row=1, column = 0 )
    Label(window, text=" Last Name").grid(row=2,column = 0)
    Label(window, text = "Age").grid(row = 3,column = 0)
    Label(window, text = "Height (m)").grid(row = 4,column = 0)
    Label(window, text = "Weight(kg)").grid(row = 5,column = 0)
    UserName = Entry(window, width = 25)
    UserName.grid(row = 0, column =1)
    User_FirstName = Entry(window, width = 25)
    User_FirstName.grid(row=1, column = 1)
    User_Surname = Entry(window, width = 25)
    User_Surname.grid(row=2, column = 1 )
    User_Age = Entry(window, width = 25)
    User_Age.grid(row = 3, column = 1)
    User_Height = Entry(window, width = 25)
    User_Height.grid(row = 4, column = 1)
    User_Weight = Entry(window, width = 25)
    User_Weight.grid(row = 5, column = 1)
    Button(window, text='Quit', command=window.quit).grid(row=10, column=0, sticky=W, padx = 100,  pady=100)
    Button(window, text='Submit', command= lambda: submit_details([ UserName.get(), User_FirstName.get(), User_Surname.get(), User_Age.get(), User_Height.get(),User_Weight.get()])).grid()

def submit_details(details):
    user_name_list = cursor.execute(''' SELECT UserName FROM GetUserDataInput''')
    if details[0] in [username[0] for username in user_name_list]:
       messagebox.showerror("Error", "User Already Exists")


    else:
        sql = '''INSERT INTO GetUserDataInput(UserName, FirstName, LastName, Age, Height, Weight)
                VALUES(?, ?, ?, ?, ?, ?)'''
        cursor.execute(sql, details)

这里，sql只是一个class，带有一些与数据库有关的方法，例如：创建连接、游标和查询。我没有把它放在这里以防它是多余的，但会很乐意把它放在需要的地方。

感谢任何能提供帮助的人。

Answer 1

最简单的解决方案是不使用 lambda。创建一个合适的函数，它会更容易编写，更容易理解，更容易调试。在这种特定情况下，lambda 只会让您的代码变得比它需要的更复杂。

在这种特定情况下，只需将调用移至 submit_details 函数内的 get。

def second_win():
    ...
    Button(window, text='Submit', command= submit_details).grid()
    ...

def submit_details():
    username = UserName.get()
    firstname = User_FirstName.get()
    surname = User_Surname.get()
    age = User_Age.get()
    height = User_Height.get()
    weight = User_weight.get()
    ...

如果您希望 submit_details 保持原样，请创建一个中间函数以在将数据传递给 submit_details 之前从 UI 获取数据：

def second_win():
    ...
    Button(window, text='Submit', command=submit).grid()
    ...

def submit():
    username = UserName.get()
    firstname = User_FirstName.get()
    surname = User_Surname.get()
    age = User_Age.get()
    height = User_Height.get()
    weight = User_weight.get()
    
    data = [username, firstname, surname, age, height, weight]
    submit_details(data)

我如何通过 lambda 函数处理异常？ [tkinter]

How can i exception handle through a lambda function? [tkinter]

tkinter

exception

messagebox