python - 如何从 Tkinter 的弹出窗口中获取值？

Question

我正在尝试创建一个包含一个 Entry 和一个 Button 的弹出窗口。我需要获取用户在该条目中输入的所有内容并将其返回到我的主窗口。我正在使用Python 3.6.0。我有以下代码(受到 this question 中的代码的启发):

from tkinter import *


class Application(Frame):    
    def __init__(self, master=None):
        super().__init__(master)

        self.server_address = None

        self.master = master
        self.master.title('Application')
        self.master.resizable(width=False, height=False)
        self.master.minsize(width=800, height=600)
        self.master.maxsize(width=800, height=600)

        self.grid()
        self.init_menubar()

    def init_menubar(self):
        menubar = Menu(self.master)
        self.master.config(menu=menubar)

        filemenu = Menu(menubar, tearoff=False)
        filemenu.add_command(label="Set remote server", command=self.call_server_config)
        filemenu.add_command(label="Set refresh delay", command=None)
        filemenu.add_separator()
        filemenu.add_command(label="Close", command=self.master.destroy)
        menubar.add_cascade(label='File', menu=filemenu)

        helpmenu = Menu(menubar, tearoff=False)
        helpmenu.add_command(label="Help", command=None)
        menubar.add_cascade(label='Help', menu=helpmenu)

    def call_server_config(self):
        popup = ServerConfig(self.master)
        self.master.wait_window(popup.top)
        print('DEBUG:', popup.value)
        self.create_widgets()


class ServerConfig(Frame):
    def __init__(self, master=None):
        super().__init__(master)

        self.top = Toplevel(master)
        self.label = Label(self.top, text="Server address")
        self.label.pack()
        self.entry = Entry(self.top)
        self.entry.pack()
        self.button = Button(self.top, text='Ok', command=self.top.destroy())

        self.value = self.entry.get()


if __name__ == '__main__':
    root = Tk()
    app = Application(master=root)
    app.mainloop()

这给了我错误:

Exception in Tkinter callback
Traceback (most recent call last):
  File "C:\Users\<...>\Python\Python36\lib\tkinter\__init__.py", line 1699, in __call__
    return self.func(*args)
  File "D:/Documents/<...>/tk-client.py", line 45, in call_server_config
    popup = ServerConfig(self)
  File "D:/Documents/<...>/tk-client.py", line 60, in __init__
    self.button = Button(self.top, text='Ok', command=self.top.destroy())
  File "C:\Users\<...>\Python\Python36\lib\tkinter\__init__.py", line 2363, in __init__
    Widget.__init__(self, master, 'button', cnf, kw)
  File "C:\Users\<...>\Python\Python36\lib\tkinter\__init__.py", line 2293, in __init__
    (widgetName, self._w) + extra + self._options(cnf))
_tkinter.TclError: bad window path name ".!application.!toplevel"

Answer 1

这段代码导致了异常:

self.button = Button(self.top, text='Ok', command=self.top.destroy())

这是因为 self.top.destroy() 在您创建按钮时首先被评估。因此，您无法将按钮放置在 self.top 上，因为它已被销毁。您需要将其更改为:

self.button = Button(self.top, text='Ok', command=self.top.destroy)

command=self.top.destroy 仅在按下按钮时执行 self.top.destroy