Help on my first code: Even or Odd Machine

Question

I've been trying to write a code in Python that tells if an integer is even or odd. I also want this code to know if the input is an integer or not, so that if the input is a string, it will say 'That is not an integer' instead of an error message. This is the code so far:

a = input() #Enter your integer in the input.

if int(a) % 2 == 0:
print('Your integer ', a, 'is even.')
elif int(a) % 2 == 1:
print('Your integer', a, 'is odd.')
else:
print('That is not an integer.')

When I enter an integer, it tells if it is even or odd. When I enter a float or string, I still get an error message. Could anyone tell me how I could rephrase the code? An explanation with it would help me learn too.

Thanks!

EDIT: Solved. Thanks to Toni Isotalo and Pulkit Kamboj for your help.

Answer 1

To check if the number is odd should be

int(a) % 2 != 0

To check if the input is an integer you could use try and except statements

try:
a = int(input()) #Enter your integer in the input.

if a % 2 == 0:
print('Your integer', a, 'is even.')
elif a % 2 != 0:
print('Your integer', a, 'is odd.')
except ValueError:
print('That is not an integer')

Answer 2

... yes please I've been getting so many notifications

Answer 3

Help on my first code: Even or Odd Machine
for that u have to apply exception handling like this:

a = input() #Enter your integer in the input.
try:
n=int(a)
except ValueError:
print('this is a string, type an integer')
else:
if n % 2 == 0:
print('Your integer ', a, 'is even.')
elif n % 2 == 1:
print('Your integer', a, 'is odd.')
else:
print('this is a floating point number')

Answer 4

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Answer 5

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Answer 6

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Answer 7

that the site I want to make py

Answer 8

toni

Answer 9

ok

Answer 10

Is there a way to download on the Samsung phone

Answer 11

because I m at work

Answer 12

yes

Answer 13

I got telegram

Answer 14

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