Python - 函数中的变量未定义

Question

我正在尝试模拟彩票，但我的功能有点问题。这是我正在尝试做的事情：

运行窗口 1()
window1() --> destroy_window1() --> window2()
window2() --> 销毁 window2 或 retry() --> window1()

当我到达 destroy_window1() 时发生错误，我收到以下消息："NameError: name 'e1' is not defined"。我该如何解决这个问题？

我听说您应该在函数外预先定义变量。所以，我尝试只输入 e1 = 1 等，但随后我收到消息："AttributeError: 'int' object has no attribute 'get'"。既然是入口，不知道怎么预定义。

from tkinter import*
import random

Part1 = list(range(1,51))
Part2 = list(range(1,11))

Numbers = [0]*7

for n in range (5):
    Number = random.choice(Part1)

    Position = Part1.index(Number)
    del Part1[Position]

    Numbers[n] = Number


for i in range (2):
    Number = random.choice(Part2)

    Position = Part2.index(Number)
    del Part2[Position]

    Numbers[5+i] = Number

print (Numbers)


def destroy_window1():
    global Guess
    Guess = [e1.get(), e2.get(), e3.get(), e4.get(), e5.get(), e6.get(), e7.get()]
    master1.destroy()
    window2()

def retry():
    master2.destroy()
    window1()

def window1():

    master1 = Tk()
    master1.title('Lottery')
    Label(master1, text="Guess numbers:").grid(row=0)

    e1 = Entry(master1, width=2)
    e2 = Entry(master1, width=2)
    e3 = Entry(master1, width=2)
    e4 = Entry(master1, width=2)
    e5 = Entry(master1, width=2)
    e6 = Entry(master1, width=2)
    e7 = Entry(master1, width=2)


    e1.grid(row=0, column=1, padx=5)
    e2.grid(row=0, column=2, padx=5)
    e3.grid(row=0, column=3, padx=5)
    e4.grid(row=0, column=4, padx=5)
    e5.grid(row=0, column=5, padx=5)
    e6.grid(row=0, column=7, padx=5)
    e7.grid(row=0, column=8, padx=5)

    master1.grid_columnconfigure(6, minsize=20) # Creates an empty column (nr. 6) with width 20

    Button(master1, text='OK', command=destroy_window1).grid(row=3, column=3, sticky=W, pady=5)



    master1.mainloop()


def window2():

    master2 = Tk()
    master2.title('Check results')
    Label(master2, text="Drawn numbers:").grid(row=0, column=0, sticky=W)
    Label(master2, text="Your numbers:").grid(row=1, column=0, sticky=W)

    for n in range (7):
        Label(master2, text=Numbers[n]).grid(row=0, column=n+1, sticky=W, padx=5)

        if str(Numbers[n]) == Guess[n]:
            Label(master2, text=Guess[n], bg="green").grid(row=1, column=n+1, sticky=W, padx=5)
        else:
            Label(master2, text=Guess[n], bg="red").grid(row=1, column=n+1, sticky=W, padx=5)

    Button(master2, text='Quit', command=master2.destroy).grid(row=3, column=3, sticky=W, pady=5)
    Button(master2, text='Retry', command=retry).grid(row=3, column=4, sticky=W, pady=5)

    master2.mainloop ()

window1()

我还不能投票，提前致谢！

Answer 1

The error occurs when I reach destroy_window1(), where i get the following message: "NameError: name 'e1' is not defined". How can I solve this problem?

问题是 destroy_window1() 函数不知道 e1 变量，因为 e1 是在 window1() 函数中定义的（而且它不是全局的).

一个简单的修复方法是将所有 e 变量放入 list 并将该 list 作为参数传递给 destroy_window1() 函数。用一个简单的 for 循环制作列表，这不仅解决了您的问题，而且还使您的代码更清晰，更易于阅读，并且将来更容易更改其功能。

像这样：

def destroy_window1(e_list):
    global Guess
    Guess = []

    for e_item in e_list:
        Guess.append(e_item.get())

    master1.destroy()
    window2()

def window1():

    master1 = Tk()
    master1.title('Lottery')
    Label(master1, text="Guess numbers:").grid(row=0)

    e_list = []

    for i in range(7):
        temp_e = e1 = Entry(master1, width=2)
        temp_e.grid(row=0, column=i, padx=5)
        e_list.append(temp_e)

    master1.grid_columnconfigure(6, minsize=20) # Creates an empty column (nr. 6) with width 20

    Button(master1, text='OK', command=lambda :destroy_window1(e_list)).grid(row=3, column=3, sticky=W, pady=5)

    master1.mainloop()

此解决方案的一部分涉及 lambda 功能。这是因为（您可能已经注意到）命令选项通常不能为函数提供参数。 lambda 函数的使用使这成为可能。（继续阅读 Lambda Functions Here）

Python - 函数中的变量未定义

Python - Variable in function not defined

tkinter

function

undefined

python-3.7