I am stuck. Please help

Question

birthDay = input (“Enter Birth Year”)
If birthDay > 1997:
Print (“you cannot drink alcohol”)
Elif birthDay < 1997:
Print (“you can drink alcohol”)

I am getting error that < > not supported between instances of str and int

Answer 1

What if I was born in 1997? (that's my case lol)

Answer 2

benjamin A more clear way of doing that:

birthYear = input("Enter birth year: ")
while not birthYear.isdigit():
birthYear = input("Please, enter a valid year: ")
birthYear = int(birthYear)

Answer 3

thank you. i ended up typing int. do i put int if i am asking for numbers?

Answer 4

yes !! int for numbers, float for decimals ^__^

Answer 5

the int func converts a string to a integer. And the input func gives you a string so you have to convert the string to int with int() func.
If the user inputs a text thats not an int an error occurs. Because the int() func can not convert something like ABC into a number

Answer 6

thank you all. i was able to run it with no issue but inhabe to change my input. i had to ask the user to choose 1 to 100

Answer 7

Mark Coching 「ＨＡＰＰＹ ＴＯ ＨＥＬＰ」 Ayesha Qudas Tortello Mr-BlackHat Tim Eduardo Petry Guys look closely there r two errors 1) input to be in int so int(input("") and 2nd Print "p" should be lower case ..it will report error ...hope u got it 😎

Answer 8

You should use >= instead of > in the first line of IF block. Like this:

If birthDay >= 1997:

Also, if you don't have more to compare, use ELSE instead of ELIF.

Answer 9

Abhay I think he is referring to the uppercase 'E' on Elif. But this would also apply to the If... and Print...

Answer 10

Abhay Singh what do you mean by exit() in function??

import sys

def exxit():
sys.exit()

val = int(input(“Press 9 to exit: “))
if val == 9:
exxit()

You can also use exit() directly

Answer 11

Abhay, have a read of https://stackoverflow.com/questions/6501121/difference-between-exit-and-sys-exit-in-JUMP_LINK__&&__python__&&__JUMP_LINK

Answer 12

You cannot compare the numbers and strings with each other. please convert you input into integer so that you can compare.
i.e. int(input("enter birth year"))

and then try

Answer 13

I would get the input as a string and then use try...except to convert it to an int.

This way, if the input cannot be converted to int, you can handle the error instead of your program stopping. Something like:

birthDay = input ("Enter birth year: ")
while True:
try:
birthDay = int (birthDay)
break
except ValueError:
birthDay = input ("Please enter a valid year: ")

There is other validation you could include such as range check, but this is a good start.

Answer 14

Eduardo, well written and you're right for this example. I was considering a scenario where you might want to catch other errors too by adding more except clauses.

Edit. I'm interested if there are better ways to do this than I suggested.

Answer 15

# use this code :-

birthDay = int(input("Enter Birth Year"))
if 1997 >= birthDay :
print("you cannot drink alcohol")
else :
print("you can drink alcohol")

# or

birthDay = input("Enter Birth Year")
if str(1997) >= birthDay :
print("you cannot drink alcohol")
else :
print("you can drink alcohol")

# both will give same output

Answer 16

and there are more solution for this

Answer 17

Inout always produces a type str that's where the error is coming from you need cast to an int for it to work

Answer 18

I'll Benjamin thankyou for clearing my dought it is giving an error and birthDay is not done in int()
it must be like this
birthDay = int(input("Enter Birth Year ")

Answer 19

Abhay no problem. The issue you will have is if the input cannot be converted to an integer. This wouuld happen if the user enters alphabetic characters instead of numeric ones.

Answer 20

Input is always taken as stings .So you need to type cast it with int