关闭前一个顶层窗口后创建顶层 tkinter 窗口答案

【问题标题】：Create Toplevel tkinter Window after the closure of previous Toplevel window关闭前一个顶层窗口后创建顶层 tkinter 窗口
【发布时间】：2020-08-14 11:06:56
【问题描述】：

我有一个包含 n 个元素的列表，我需要为每个元素创建一个弹出菜单。每个弹出窗口都会包含一些复选框。

条件：新的Toplevel弹出窗口必须在前一个Toplevel窗口关闭后打开，而不是同时打开

我的代码：

from tkinter import *

#root gui
root = Tk()
root.title("test")
root.geometry("300x400")

# Need three popup windows
a = ["one", "two", "three"]
b = []

def open():
    for _ in range(len(a)):
        top = Toplevel()
        top.title("selections")
        def next_window():
            top.destroy()
            show() # This function is supposed to show the selections of each popup window on root gui
        for i in range(3):
            b.append(IntVar())
            b[i].set(0) # de selecting all checkboxes intiially
            # checkboxes
            Checkbutton(top, text=a[i], variable=b[i]).pack()
        Button(top, text = "Submit", command=next_window).pack()
        Button(top, text = "skip", command=top.destroy).pack() # this button is used to skip the popup if no selection required

def show():
 # printing selections made on each popup window
 for i in range(3):
  Label(root, text=b[i].get()).pack()

mB = Button(root, text="print selections", command=open).pack()

root.mainloop()

我的担忧：我现在正在同时打开所有三个弹出窗口。

【问题讨论】：

标签： python user-interface tkinter toplevel

【解决方案1】：

你需要在for循环的最后调用top.wait_window()：

for _ in range(len(a)):
    top = Toplevel(root)
    top.title("selections")
    def next_window():
        top.destroy()
        show() # This function is supposed to show the selections of each popup window on root gui
    for i in range(3):
        b.append(IntVar())
        b[i].set(0) # de selecting all checkboxes intiially
        # checkboxes
        Checkbutton(top, text=a[i], variable=b[i]).pack()
    Button(top, text = "Submit", command=next_window).pack()
    Button(top, text = "skip", command=top.destroy).pack() # this button is used to skip the popup if no selection required
    top.grab_set() # route all events to this window
    top.wait_window() # wait for current window to close

【讨论】：