Python Closing Toplevel Window Error

Question

I wanted this code to create a popup error window that destroys itself after 4 seconds but can also be closed via a button.

def error(self):
    top = Toplevel()
    top.after(4000, lambda: top.destroy())
    center_window(300,100, top)
    top.title("Error")
    Label(top, text="Please enter a valid code", height=3, width=200).pack()
    ok = Button(top, text="OK", command=top.destroy)
    ok.pack()
    ok.bind("<Return>", lambda a: top.destroy())
    ok.focus_set()

I have run the code and it works fine 90% of the time except sometimes it throws this error:

 TypeError: <lambda>() takes exactly 1 argument (0 given)

I have done research that says it is Tkinters threading. I am not sure if this is my issue but when I take out this line of code:

top.after(4000, lambda: top.destroy())

It seems to work. If anyone could help me, I have taught myself what I know of python, so I am sure there are holes in my learning. I think I may need to somehow use the main thread of execution to destroy this window, or otherwise create my own custom widget. Any help is appreciated.

This has absolutely nothing to do with threading. Tkinter runs in a single thread at all times. — Bryan Oakley, Sep 12 '17 at 10:56

score 1 · Answer 1 · answered Sep 12 '17 at 08:03

1

When using after or bind, you don't need to use lambda. Instead, for example, use:

top.after(4000, top.destroy)

which references the function top.destroy directly.

answered Sep 12 '17 at 08:03

Temsia Carrolla

191
7

score 0 · Answer 2 · answered Sep 12 '17 at 08:05

You can directly bind the function to be called instead of using a lambda:

    top.after(4000, top.destroy)
...
    ok.bind("<Return>", top.destroy)

You would use a lambda, for instance, if you needed to pass arguments to the function; this is not the case here.

Python Closing Toplevel Window Error

2 Answers2