ValueError: invalid literal for int() with base 10: for any string

Question

def sm():
    seconds_entry_get = seconds_entry.get()
    if seconds_entry_get == type(str):
        messagebox.showerror("Wrong Input", "Please type a number or decimal")
    else:
        converted = int(seconds_entry_get) / 60
        conversion_text.delete("1.0", "end")
        conversion_text.insert("1.0", converted)

There is some "invalid literal" error and I am having trouble fixing it.

what happens if you print(seconds_entry_get) at the start of your else statement? Does it look like a number? — Chris
– Chris, Commented Jun 15, 2021 at 17:41
if seconds_entry_get == type(str): does not do what you think it does, I assume what you were trying to do was if isinstance(seconds_entry_get, str): — Cory Kramer
– Cory Kramer, Commented Jun 15, 2021 at 17:41
This is quite wrong because <Entry>.get() always returns a str. — Delrius Euphoria
– Delrius Euphoria, Commented Jun 15, 2021 at 18:04

Cory Kramer · Accepted Answer · 2021-06-15 17:43:24Z

1

If you want to convert to int but handle invalid inputs I would instead use try and except

def sm():
    seconds_entry_get = seconds_entry.get()
    try:
        converted = int(seconds_entry_get) / 60
        conversion_text.delete("1.0", "end")
        conversion_text.insert("1.0", converted)
    except ValueError:
        messagebox.showerror("Wrong Input", "Please type a number or decimal")

answered Jun 15, 2021 at 17:43

Cory Kramer

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