我有一个n个元素的列表,我需要为每个元素创建一个弹出菜单。每个弹出窗口都包含一些复选框。
条件:新的Toplevel弹出窗口必须在关闭以前的Toplevel窗口后打开,而不是同时打开所有窗口
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(),My concern,:这三个弹出窗口现在都同时为我打开。

发布于 2020-08-14 11:23:09
您需要在for循环的末尾调用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 closehttps://stackoverflow.com/questions/63411575
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