Question

我一直希望在tkinter中创建一个弹出窗口小部件，用户可以在其中输入一些数据。但是在测试时，我想到了以下错误：

  File "returnWeeksGUI_v1.py", line 363, in __init__
    self.content = program.popup("Add_label")
  File "returnWeeksGUI_v1.py", line 296, in popup
    self.w=popupWindow(self.master,popup_type)
  File "returnWeeksGUI_v1.py", line 42, in __init__
    self.bu=tk.Button(top,text='Ok',command=self.cleanup(win_type))
  File "C:\Users\Fernanda\AppData\Local\Programs\Python\Python37-32\lib\tkinter\__init__.py", line 2369, in __init__
    Widget.__init__(self, master, 'button', cnf, kw)
  File "C:\Users\Fernanda\AppData\Local\Programs\Python\Python37-32\lib\tkinter\__init__.py", line 2299, in __init__
    (widgetName, self._w) + extra + self._options(cnf))
_tkinter.TclError: bad window path name ".!toplevel"

已经进行了3天的调试和编辑代码，但仍然没有找到合适的解决方案。

代码

我有一些导入和变量：

import tkinter as tk

models = {
    "text":["TXT",50,50],
    "DD":["DD",50,50],
    "MM":["MM",50,50],
    "AA":["AA",50,50],
    "MC":["MC",50,50],
    "DP":["DP",50,50],  
    }

我有一个tkinter课：

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

        #This class has some variables:
        self.pu= program.popup("Add_label")

    def popup(self,popup_type):
            self.w=popupWindow(self.master,popup_type)
            self.master.wait_window(self.w.top)

我还有一个从a question in StackOverflow获取的弹出代码，出于我的目的对其进行了修改。它显示了用户必须完成/选择的一些单选按钮/条目。

class popupWindow(object):
    def __init__(self,master,win_type):
        top=self.top=tk.Toplevel(master)
        if str(win_type) == "Add_label":
            """
            Here goes a large piece of code that is irrelevant.  
            """
            # Until I have this set of statements which break the program.
            self.bu=tk.Button(top,text='Ok',command=self.cleanup(win_type))
            self.bu.pack()

        elif str(win_type) == "Add_box":
            #...
            self.bu=tk.Button(top,text='Ok',command=self.cleanup(win_type))
            self.bu.pack()

        elif str(win_type) == "Delete_label":
            #...
            self.bu=tk.Button(top,text='Ok',command=self.cleanup(win_type))
            self.bu.pack()

        elif str(win_type) == "Delete_box": 
            #...
            self.bu=tk.Button(top,text='Ok',command=self.cleanup(win_type))
            self.bu.pack()

    def cleanup(self,win_type):
        #...
        # At the end, a destroy method is called.
        self.top.destroy()

实例如下：

root = tk.Tk()
program = App(root)
program.mainloop()

弹出式菜单不是由__init__()函数中的变量触发的。实际上，当按下某个按钮时，它会显示出来。但是，在这里我将其替换为变量，因为它更易于分析。触发弹出窗口的按钮很好。我已经检查过了。

如果有人可以提供帮助，我将不胜感激。谢谢。

Answer 1

您现在在创建按钮期间正在调用self.cleanup（win_type），从而破坏了您要向其添加按钮的窗口。一种解决方案是将其写为command = lambda：self.cleanup（win_type），以便将调用推迟到实际单击按钮之前。 –杰森哈珀的评论答案

Python 3.x _tkinter.TclError：错误的窗口路径名“。！toplevel”

1 个答案: