CODEWITHSUNDEEP
exceptionpython-3.x
Mike
Mike

Reputation: 195

exception in input in python

When running the following code:

try:
   key=int(input())
except ValueError as string:
   print("Error is within:",string)

for example, if one puts 'rrr' this exception will rise, since 'rrr' does not support (int)

However, instead ot putting the actual string, it puts: "invalid literal for int() with base 10: 'rrr' "

How do I make it work so that the variable 'string' actually gets the wrong input that the user gave (in this example, I want it to print: 'Error is within: rrr')

Thanks a lot

Upvotes: 0

Views: 1541

Answers (3)

Fabricator
Fabricator

Reputation: 12772

You can just parse the error msg :D

print("Error is within:", string.args[0][41:-1])

Upvotes: 0

Bryan
Bryan

Reputation: 2068

Your issue comes from the fact that the variable string is the error message for a ValueError exception. If you wanted to print out the user's invalid input, you would need to create a variable that stores the user's input before your try/except. For example:

userInput = input()
try:
   key=int(userInput)
except ValueError:
   print("Error is within:",userInput)

Upvotes: 5

Eevee
Eevee

Reputation: 48546

Store the input and convert it to an int separately.

key_str = input()
try:
    key = int(key_str)
except ValueError:
    print("Error is within:", key_str)

Upvotes: 5

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