从背景python脚本打开弹出窗口

Question

我正在尝试使用tkinter从我的代码打开弹出gui窗口，该代码使用crontab在后台运行，但是它会出错。 当我从控制台运行此脚本时，它工作正常。 相关部分：

def copy():
  global threshold
  global primary
  global secondary

  root = Tk()
  copy_popup = Toplevel(root)
  copy_label = Label(copy_popup, text="Copying Files...")
  copy_label.grid(column=0, row=0)

  progress = ttk.Progressbar(copy_popup, orient="horizontal", length=200, mode="indeterminate")
  progress.grid(column=1, row=0)

  root.after(1, start_copy, progress, root)
  root.mainloop()


def start_copy(progress, root):
    copy_thread = threading.Thread(target=copy_files)
    print("got here")
    progress.start()
    copy_thread.start()
    copy_thread.join()
    progress.stop()
    root.destroy()

错误追溯：

    Traceback (most recent call last):
  File "/home/nimrco/PycharmProjects/Cron/cron.py", line 248, in <module>
    main()
  File "/home/nimrco/PycharmProjects/Cron/cron.py", line 230, in main
    pending()
  File "/home/nimrco/PycharmProjects/Cron/cron.py", line 81, in pending
    copy()
  File "/home/nimrco/PycharmProjects/Cron/cron.py", line 154, in copy
    root = Tk()
  File "/usr/lib/python3.5/tkinter/__init__.py", line 1880, in __init__
    self.tk = _tkinter.create(screenName, baseName, className, interactive, wantobjects, useTk, sync, use)
_tkinter.TclError: no display name and no $DISPLAY environment variable

任何人都有这方面的解决方案吗？提前谢谢你。