How can I check if a string is convertable to integer without isdigit or similar?-CodePudding

I'm working on a project for university, and I need to use an integer from a user input. However I'm trying to make sure my code doesn't break as soon as someone types something the code wasn't expecting, a letter or word instead of a number for example.

After asking, I was told I'm not to use isdigit() or similar functions. Is there a way out of this or should I ignore this weak point in my code?

CodePudding user response：

Technically this doesn't use any functions like isdigit()...

all(c in "0123456789" for c in string)

Examples:

>>> string = "239a0932"
>>> all(c in "0123456789" for c in string)
False
>>> string = "9390239"
>>> all(c in "0123456789" for c in string)
True

CodePudding user response：

You can either just try to convert the input to an int, and if it fails with an exception, it wasn't a number. Or you use a regular expression.

import re

entered = input("Enter a text: ")

# Check as regular expression
pattern = re.compile(r"^\d $")
if pattern.match(entered):
    print("re says it's a number")
else:
    print("re says it's not a number")

# Try to convert
try:
    asNum = int(entered)
    print("can be converted to a number")
except ValueError:
    print("cannot be converted to a number")

CodePudding user response：

Use try/except:

try:
    num = int(input("Enter a number"))
except ValueError:
    # do whatever you want to do if it wasn't a valid number

If you want to re-prompt the user until they enter a number, and then go ahead with whatever you needed num for, that looks like:

while True:
    try:
        num = int(input("Enter a number"))
        break  # end the loop!
    except ValueError:
        print("nope, try again!")
        # loop continues because we didn't break it

print(f"{num}   {num} = {num * 2}!")

CodePudding user response：

For example:

if len(my_probably_digits_str) == len("".join([x for x in my_probably_digits_str if x in "0123456789"])):
    print("It's convertable!")