Variable syntax error (Python)

Question

I'm new to coding with python and I've came across an issue with my variables in an "if" or "else" command.

Answer = input("Yo stupid, What\'s nine plus ten?")
if Answer == 19: print("Your smarter then me.")
else Answer >= 21: print("You stupid!")

^^ When I attempt to run my code It tells me I have made an error with my syntax. It always pops up with my variable "Answer" in the "if" or "else" command. Any help would be greatly appreciated. Thanks in advance.

Additionally, input returns a string, so "19" == 19 will be false. — Mark Tolonen
– Mark Tolonen, Commented Oct 20, 2016 at 0:58

TerryA · Accepted Answer · 2016-10-20 00:56:00Z

4

elif (short for "else if") should be used if you want to test another condition for Answer. else is used on its on for every other option that Answer could be that doesn't fit the previous conditionals.

answered Oct 20, 2016 at 0:56

TerryA

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3 Comments

Mangohero1 Over a year ago

Should also mention that the conditionals are resulting in a TypeError, it should be if Answer == '19', elif Answer >= '21' , but hit the nail on the head with the first one :-)

Oscar Cheung Over a year ago

Thanks for helping, however, I tried what you said (to the best my pea-brain can comprehend), but now it says I have an error with "print" Sorry if i'm a bit of a nuisance here.

Answer = input("Yo stupid, What\'s nine plus ten?") if Answer == '19' print("Your smarter then me.") elif Answer >= '21' print("You stupid!")

Mangohero1 Over a year ago

Don't forget to use the colons at the end of each if/elif

chepner · Accepted Answer · 2016-10-20 01:05:52Z

0

else: does not take a condition afterwards. In your case use elif <condition>:.

edited Oct 20, 2016 at 1:05

chepner

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answered Oct 20, 2016 at 0:57

Tiphaine

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