python - 创建一个带有未知数量复选框的窗口 - Python/tkinter

Question

我正在为我的计算机科学类(class)开发一个涉及 python 和 tkinter 的项目。我正在尝试制作一款功能齐全的大富翁游戏，而且进展顺利。我终于遇到了一个我似乎无法克服的障碍。我正在尝试制作一个界面来抵押用户的属性(property)，我想使用 tkinter checkbuttons 来获取用户输入，然后抵押所有已检查的属性(property)。这是我制作的类(class)的片段:

from tkinter import *

class Mortgager(Tk):
    def __init__(self,coorder):    # 'coorder' is a class that coordinates all of the
        self.coorder = coorder     # other classes together

        Tk.__init__(self,className='Mortgager')
        self.title('Mortgaging')

        self.cbuttons = []
        self.intvars = []
        for prop in coorder.active_player.properties:    # iterate through player's currently owned properties
            if not prop.mortgaged:
                self.intvars.append(IntVar(self,0))
                self.cbuttons.append(Checkbutton(self,
                                                 variable=self.intvars[-1],text=prop.get_name(),
                                                 # Most recent intvar, method returns name of property

                                                 command=self.update_cash_onscreen)
                                                 #### Not sure what to do here...
                self.cbuttons[-1].var = self.intvars[-1]
                self.cbuttons[-1].text = prop.get_name()
        i = 0
        for cbutton in self.cbuttons:    # Every three properties, new column
            cbutton.grid(column=i//3,row=i%3,sticky=W,
                         padx=5,pady=5)
            i += 1
        # Haven't finished the rest of the class...

我的问题是这样的:如何创建任意数量的复选按钮，然后告诉“即时”单击了哪些复选按钮，更新某种显示当前数量的Label抵押，用 StringVar 或类似的东西，然后用总金额做一些事情？

提前致谢!

Answer 1

我完全不理解你的代码，但是如果你想在“ctrls”列表中创建带有标签的 N 个检查按钮，请尝试这个

# if ctrls is a list of all lables to your checkboxes
# i is the count and j is the text of label
for i,j in enumerate(ctrls): #what ever loop you want
    var = IntVar()
    c = Checkbutton(self.master,text=j,variable=var)
    boxes.append([j.strip(),var,c])

稍后如果您想检查哪些按钮被选中

for i in boxes:
    if i[1].get()==0:
        #do what ever you want 
        i[2].destroy()