Question

我创建了一个直接从根窗口继承的简单类，它在Mac机器上运行完美。但是，当我尝试在我自己的基于Ubuntu的机器上运行它时，它崩溃并出现以下错误。造成这种情况的原因是什么？

Traceback (most recent call last):
  File "/home/estilen/Dropbox/Python/email_viewer/dialog.py", line 15, in <module>
    Dialog('This is definitely working')
  File "/home/estilen/Dropbox/Python/email_viewer/dialog.py", line 7, in __init__
    tk.Tk.__init__(self, dialog_message)
  File "/usr/lib/python3.5/tkinter/__init__.py", line 1871, in __init__
    self.tk = _tkinter.create(screenName, baseName, className, interactive, wantobjects, useTk, sync, use)
_tkinter.TclError: couldn't connect to display "This is definitely working"

代码：

import tkinter as tk


class Dialog(tk.Tk):

    def __init__(self, dialog_message):
        tk.Tk.__init__(self, dialog_message)
        self.geometry('400x100')
        label = tk.Label(self, text=dialog_message)
        label.pack(anchor='center', side='top', pady=10)
        self.mainloop()


if __name__ == '__main__':
    Dialog('This is definitely working')

Answer 1

tk.Tk.__init__(self, dialog_message)

这对我来说并不合适。 Tk构造函数的第一个位置参数是screenName。如果您不想在名为＆＃34;的显示屏上显示您的窗口，这肯定有效，那么就不要提供该论点。

tk.Tk.__init__(self)

...或者，更具有惯用力，

super().__init__()

Tkinter继承了在Ubuntu上运行不正常的类

1 个答案: