How to repeat try-except block

Question

I have a try-except block in Python 3.3, and I want it to run indefinitely.

try:
    imp = int(input(\"Importance:\\n\\t1: High\\n\\t2: Normal\\n\\t3: Low\")         


        

Answer 1

Put it inside a while loop and break out when you've got the input you expect. It's probably best to keep all code dependant on imp in the try as below, or set a default value for it to prevent NameError's further down.

while True:
  try:
    imp = int(input("Importance:\n\t1: High\n\t2: Normal\n\t3: Low"))

    # ... Do stuff dependant on "imp"

    break # Only triggered if input is valid...
  except ValueError:
    print("Error: Invalid number")

EDIT: user2678074 makes the valid point that this could make debugging difficult as it could get stuck in an infinite loop.

I would make two suggestions to resolve this - firstly use a for loop with a defined number of retries. Secondly, place the above in a function, so it's kept separate from the rest of your application logic and the error is isolated within the scope of that function:

def safeIntegerInput( num_retries = 3 ):
    for attempt_no in range(num_retries):
        try:
            return int(input("Importance:\n\t1: High\n\t2: Normal\n\t3: Low"))
        except ValueError as error:
            if attempt_no < (num_retries - 1):
                print("Error: Invalid number")
            else:
                raise error

With that in place, you can have a try/except outside of the function call and it'll only through if you go beyond the max number of retries.