Error when Executing a python code

Question

I am new to Python programming. I am writing the code below, and when I execute it the IDE returns an error message : TypeError: unorderable types: str() < int()

Code below :

print("What is your name?")


name = input()

print("What is your age?")

age = input()

if name=='Jack':

  print ("Hello Jack")

elif age<12:

    print("You are not Jack")

The error

    elif age<12:
TypeError: unorderable types: str() < int()

It seems that you are comparing a string with an integer. – Mohit Sharma Nov 01 '16 at 21:33 — Mohit Sharma, Nov 01 '16 at 21:33
You need to cast your age input to an int. – idjaw Nov 01 '16 at 21:35 — idjaw, Nov 01 '16 at 21:35
So many duplicates of this, but where to find them? – Nov 01 '16 at 21:35 — , Nov 01 '16 at 21:35

score 1 · Answer 1 · answered Nov 01 '16 at 21:39

Tip:

print('something')
input()
# same as
input('something')

Then, input returns in python 3 a string. And you cannot compare a string with an int.

It's like if you were doing '5' < 2. You need to transform '5' into an int. And it's pretty easy: int('5') == 5

name = input("What is your name?")

age = input("What is your age?")

if name == 'Jack':
     print("Hello Jack")

elif int(age) < 12:

    print("You are not Jack")

Matt

score 0 · Accepted Answer · answered Nov 01 '16 at 21:38

0

input() returns a string. You cannot directly compare a string to an integer.

Convert age to an integer by calling int():

age = int(input())

answered Nov 01 '16 at 21:38

John Gordon

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Error when Executing a python code

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