How to hide default error message(Traceback) in python

Question

I wanted to know how to hide errors in python: Let's say I made a calculator and then typed:

if number1 != int:
     print("NaN")

But it will print this message and give out the error which is built into python which is: Traceback (most recent call last):

but how do I hide this Traceback error and only show the error message which is "NaN"

Thank you for the answer.

if number1 != int: makes no sense. What exactly are you trying to do? — user15801675
– user15801675, Commented Aug 2, 2021 at 13:24

user15801675 · Accepted Answer · 2021-08-02 13:29:24Z

2

Even though you are talking about try...except, the following statement makes no sense.

You telling python to compare 1 to data type int

if number1 != int: print("NaN")

If you want to check a particular data type, use isinstance(<variable>,<data type>)

if isinstance(number1,int): print("NaN")

You can use try...except method to catch various errors:

try:
    number1=int(input("Enter a number: "))
    ...
except ValueError:
    print("NaN")

Note that this will catch only ValueError. If you want to catch all errors, use except Exception or except:

answered Aug 2, 2021 at 13:29

user15801675

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2 Comments

Ohk thank you so much! Sorry for the incorrect meaning in the code.

@kronox, no problem

kaiserm99 · Accepted Answer · 2021-08-02 13:28:47Z

0

To "hide" a Error message (NameError) you can try the following, but this is only because number1 is not defined:

try:
    if number1 != int: print("NaN")

except NameError:
    print("Error is hidden!")

except:
    print("Catch all other Exceptions!")

For more see the following link.

I think you want to check if the inserted numer is not an integer. This can be done with the following code:

number = "asdf"

if type(number) != int: 
    print("NaN")

else:
    print("Valid number!")

answered Aug 2, 2021 at 13:23

kaiserm99

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