Simple Input and Conditions problem

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January 1st, 2013, 11:41 PM

sijones

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Simple Input and Conditions problem

I'm trying to do this

age = input("Pleas enter your age")
if age> 60:
Print ("Wow you are quite old")
else:Print ("are not that old")

i get an error

Traceback (most recent call last):
File "/Users/shallahjones/Desktop/python test.py", line 2, in <module>
if age> 60:
TypeError: unorderable types: str() > int()
>>>

why is it not printing correctly. Im new to this.

January 1st, 2013, 11:44 PM

sijones

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January 1st, 2013, 11:59 PM

sijones

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January 2nd, 2013, 01:15 AM

Nyktos

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The behaviour of the input function was changed from Python 2 to Python 3. In Python 2, it would evaluate the input given to it; while in Python 3 it simply returns it as a string. If you want an int, you'll have to call int() (or eval(), but don't do that) on it, like so:

Code:

age = int(input("Please enter your age: "))
if age > 60:
    print("Wow, you are quite old.")
else:
    print("You are not that old.")

In addition, please put your code in code tags in the future. Code tags will preserve indentation which is very important in Python.

Comments on this post

Dietrich agrees!

January 2nd, 2013, 08:25 AM

b49P23TIvg

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print not Print nor PRINT

__________________
[code]Code tags[/code] are essential for python code!

January 14th, 2013, 04:22 PM

merdox

Registered User

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try this:

Code:


age = int(input("Please enter your age:\n ")) # \n will put your input in new line
if int(age) > 60: # you have to compare an integer with integer !
    print("Wow, you are quite old.")
else:
    print("You are not that old.")

Simple Input and Conditions problem

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