How do you use an if statment to only except integers and give an invalid entry message?

Question

I'm making a simple guessing game in python and was trying to create an "Invalid entry" message for when the user enters in any input that is not an integer.

I have tried to use just 'int' in an if statement to address all integers, but that is not working. I know that I have the syntax wrong. I'm just not sure what the correct syntax to do it would be.

import random
play = True

while play:
    count = 1

    hidden = random.randrange(1,5)


    guess = int(input("Guess a number between 1 and 5:"))

    if guess != int
        guess = int(input("Invalid Entry. Please enter an Integer between 1 and 5:"))


    while guess != hidden:
        count+=1
        if guess > hidden + 10:
            print("your guess is to high!")
        elif guess < hidden -10:
            print("your too low!")
        elif guess > hidden:
            print("your really warm, but still to high!")
        elif guess < hidden:
            print("your really warm, but still to low")

        print("You have guessed incorrectly, Try again!. \n")
        #reset the guess variable and make another guess
        guess = int(input("Guess a number between 1 and 5:"))

    print("Nice!!! Your guess was correct!\n you got the correct number in" , count , "tries.")

    count = 1


    playagain = str(input("Do you want to play again?\nType yes or no: "))
    if playagain == "no" or "n" or "N" or "no thank you":
        play = False

    elif playagain == "yes" or "y" or "Y" or "YES" or "yes":
        play = True

    else: playagain != "yes" or "y" or "Y" or "YES" or "yes" "no" or "n" or "N" or "no thank you"
    playagain = str(input("Invalid Entry. Please Type yes or no: "))

This is the error that I'm getting. There may be some other mistakes in my code as well.

File "comrandomguess.py", line 18
    if guess != int
                  ^
SyntaxError: invalid syntax

Answer 1

If you really want to verify that the user entry is an int, you want to keep the input in string form. Then write a small function to test the input. Here, I'll use a list comprehension and the string join and isdigit methods, to ensure the user has only entered digits 0-9 in the string, i.e. then this function returns True (else False) (*modified as per Jack Taylor comment below, also for s = '' case):

def testForInt(s):
    if s:
        try:
            _ = s.encode('ascii')
        except UnicodeEncodeError:
            return False
        test = ''.join([x for x in s if x.isdigit()])
        return (test == s)
    else:
        return False

If you want to sandbox the user entirely, wrap it in a loop like this:

acceptable = False
while not acceptable:
    entry = input("Enter an int: ")
    if testForInt(entry):
        entry = int(entry)
        acceptable = True
    else:
        print("Invalid Entry")

If you want a simpler version with no function call(see Jack Taylor comment), this works too:

acceptable = False
while not acceptable:
    entry = input("Enter an int: ")
    try:
        entry = int(entry)
        acceptable = True
    except ValueError as e:
        print(f"Failed due to {str(e)}")

Now you've got what you know is an int, with no worries. This kind of approach to verifying user entry saves many headaches if consistently implemented. See SQL injection etc.

Answer 2

I always use this method to check if something is not an integer:

Python 3

if not round(guess) == guess: print("Do Stuff")

Python 2

if not round(guess) == guess: print "Do Stuff"

Answer 3

You need to do something like this:

play = True

while play:
    guess = input("Guess a number between 1 and 5: ")

    try:
        number = int(guess)
    except ValueError:
        print("You need to input an integer.")
        continue

    if number < 1 or number > 5:
        print("You need to input an integer between 1 and 5.")
        continue

    # ...

    print("Your number was: " + guess)
    play = False

When you first use input(), you get a string back. If you try to turn that string into an integer straight away by doing int(input()), and if the player types a string like "abcd", then Python will raise an exception.

>>> int(input("Guess a number: "))
Guess a number: abcd
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: invalid literal for int() with base 10: 'abcd'

To avoid this, you have to handle the exception by doing int(guess) inside a try/except block.

The continue statement skips back to the start of the while loop, so if you use it you can get away with only having to ask for input once.

Answer 4

Parse the user input as string to avoid ValueError.

guess = input("Guess a number between 1 and 5: ")

while not guess.isdigit() or int(guess) > 5 or int(guess) < 1: 
    guess = input("Invalid Entry. Please enter an Integer between 1 and 5: ")

guess = int(guess)

Above code ensures that user input is a positive integer and between 1 and 5. Next, convert the user input to integer for further use.

Additionally, if you want to check the data type of a python object/variable then use the isinstance method. Example:

a = 2
isinstance(a, int)

Output:
>>> True