如何定义通过另一个函数中的 for 循环创建的全局变量

Question

我收到错误

名称错误： 未定义全局名称"spinbox_0">

当我尝试从我的一个 Tkinter 旋转框中获取值并将其打印出来时。旋转框位于另一个函数中，如果可能的话，我会定义它，但是在这种情况下我不知道如何定义它，因为旋转框的名称和数量是由 for 循环创建的，具体取决于主题列表的长度。

我认为只发布代码进行调试会更容易。按下Button two按钮时会显示错误代码。

目前，我只想检索该spinbox_0值并打印出来。之后，我将使用另一个循环获取所有旋转框值。

from Tkinter import Tk, Button, Spinbox, Label
## Create a window
window = Tk()
## Give the window a title
window.title('Tester')
names = ['Name 1','Name 2','Name 3','Name 4','Name 5','Name 6','Name 7']
## Define the actions for when button one is pushed
def button1():
    print "Button one pushed"
## Create the Spinboxes for each name in the list
    spinbox_list = []
    spinbox_grid_list = []
    for number, name in enumerate(names):
        spinbox = 'spinbox_' + str(number) + ' = Spinbox(window, width = 3, from_=0, to=10)'
        spinbox_grid = 'spinbox_' + str(number) + '.grid(padx = 0, pady = 0, row = 13, column = '+ str(number) +')'
        spinbox_list.append(spinbox)
        spinbox_grid_list.append(spinbox_grid)
        spinbox_option = {'spinbox_{}'.format(i):e for i, e in enumerate(spinbox_list)}
    spinbox_option_string = 'n'.join(spinbox_option.values())
    spinbox_grid_option = {'spinbox_grid_{}'.format(i):e for i, e in enumerate(spinbox_grid_list)}
    spinbox_grid_option_string = 'n'.join(spinbox_grid_option.values())
    exec(spinbox_option_string)
    exec(spinbox_grid_option_string)
## Create a second button
    button_two = Button(window, text = 'Button two', command = button2, width = 20)
    button_two.grid(pady = 10, padx = 2, row = 14, columnspan = 9, sticky = "S") 
## Define the actions for when button two is pushed (print the spinbox values)
def button2():
    print 'spinbox_0 = ' + (spinbox_0.get())
    ## spinbox_0 is Just an example. I need all spinboxes to print out their
    ## values here somehow
button_one = Button(window, text = 'Button one', command = button1, width = 20)
button_one.grid(pady = 2, padx = 2, row = 1, columnspan = 9, sticky = 'S')
window.mainloop()

Answer 1

回避全局变量是一个坏主意™的事实，这里有一种方法可以做到这一点。创建全局变量，然后不要在函数中重新分配它。现在，您可以在button2函数中循环访问它。由于您需要旋转框对象，而不是其描述，因此请保存该对象。

*顺便说一句，通过避免exec，您可以大大简化button1功能。

spinboxes = []

# Define the actions for when button one is pushed
def button1():
    print "Button one pushed"
    # Create the Spinboxes for each name in the list
    spinbox_grid_list = []
    for number, name in enumerate(names):
        spinboxes.append(Spinbox(window, width=3, from_=0, to=10))
        spinboxes[number].grid(padx=0, pady=0, row=13, column=number)
    button_two = Button(window, text='Button two', command=button2, width=20)
    button_two.grid(pady=10, padx=2, row=14, columnspan=9, sticky="S")
# Define the actions for when button two is pushed (print the spinbox values)
def button2():
    for i, spinbox in enumerate(spinboxes):
        print 'spinbox_{} = '.format(i) + (spinbox.get())

我假设button1只被调用一次。否则，每次按下时，都会不断向全局列表中添加新的spinbox。要解决这个问题：

def button1():
    print "Button one pushed"
    global spinboxes
    spinboxes = []
    ...

我们使用 global 关键字让我们重新分配全局对象，每次调用函数时将其清除。