What should i do to print my own custom exceptions in python?

Question

I have a simple code which raises an Exception "Invalid Name"

But the output I'm getting is:

Traceback (most recent call last): File "..\Playground\", line 2, in <module> raise NameError("Invalid name!") NameError: Invalid name!
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What should i do to get out like this only without that Traceback thing and file details:

Expected output:

NameError: Invalid name!

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You see in simple words i don't wanna
Traceback (most recent call last): File "......

Is there any way? To do it?

https://code.sololearn.com/chF9udu0WpsI/?ref=app

Answer 1

https://code.sololearn.com/c7s72VWhU4QS/?ref=app

Answer 2

Sofia, FERZ
please don't spam here 🙄
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Gordon That answer was really so much helpful thank you buddy 👍😆

Answer 3

Can u say eloborate

Answer 4

name = ("123")
print("NameError:"" Invalid name!") This Is

Answer 5

You need to put the output option un the middle of name and name error

Answer 6

try:
inp = int(input())
except:
print('Please enter integer value: ' ) # just printing this when exception occurs useful in most case

finally:
print('This prints no matter if exception occured or not.

There is raise:
# one to for raising exception.

Answer 7

Yep

Answer 8

Yep

Answer 9

Hello

Answer 10

Auuuuuu😊😊😊😊😊

Answer 11

Are you in contact