CODEWITHSUNDEEP
python
Magical Man
Magical Man

Reputation: 19

Beginner: Need to find better way to get something done

I am working out of the Learn Python the Hard Way book and I have a very beginner question. I am trying to get it so that the code below does not have to go through such a weird way to check if a number was given, also, I want numbers that don't have 1 or 0 to be valid.

def gold_room():
    print("The room is full of gold, how much do you take?")

    choice = raw_input("> ")
    if "0" in choice or "1" in choice:
        how_much = int(choice)

The idea that I had was to get it to check if choice was an integer, and if so then trigger the code, but I can not figure out how to do this. I am sure the answer is very simple, and I'm sorry for such a basic question. Thank you for your help!

Upvotes: 0

Views: 64

Answers (2)

DYZ
DYZ

Reputation: 57033

If you are looking only for non-negative integers, than choice.isdigit() will suffice. It returns True or False, depending on whether choice consists of all digits or not. Complete solution:

def gold_room():
  choice = raw_input("The room is full of gold, how much do you take? ")
  if choice.isdigit():
    return int(choice)
  else:
    return None

Upvotes: 2

Chiheb Nexus
Chiheb Nexus

Reputation: 9257

You can use, also, try .. accept and catch the error. You can follow this example:

def gold_room():
    print("The room is full of gold, how much do you take?")

    choice = raw_input("> ")
    try:
        return int(choice)
    except ValueError:
        pass

print gold_room()

Output:

if for example the input was 12

>> 12

if the input isn't an integer so the output will be None:

>> None

Upvotes: 1

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