How to fix my if, elif, else statement in python?

Question

I am working on an omegle simulator for fun, where it asks your age and if you're old enough asks you if you have kik. It works fine if your age is 16 or more but if you say any less than that, it comes up with an error. This is the code:

age = input("age?\n")
if age == "1":
    print ("Too young bby")
elif age == "2":
    print ("Too young bby")
elif age == "3":
    print ("Too young bby")
elif age == "4":
    print ("Too young bby")
elif age == "5":
    print ("Too young bby")
elif age == "6":
    print ("Too young bby")
elif age == "7":
    print ("Too young bby")
elif age == "8":
    print ("Too young bby")
elif age == "9":
    print ("Too young bby")
elif age == "10":
    print ("Too young bby")
elif age == "11":
    print ("Too young bby")
elif age == "12":
    print ("Too young bby")
elif age == "13":
    print ("Too young bby")
elif age == "14":
    print ("Too young bby")
elif age == "15":
    print ("Too young bby")
else:
    kik = input("Do you have kik?\n")
yes = "yes"
if kik == yes:
    print ("add me bby")
else:
    print ("bye")

The error that comes up is:

 Traceback (most recent call last):
  File "C:/Users/Public/Documents/python/omegle.py", line 36, in <module>
    if kik == yes:
 NameError: name 'kik' is not defined

Does anyone know how to fix it?

Answer 1

The problem is that you only set kik in this block:

else:
    kik = input("Do you have kik?\n")

If this block isn't reached, kik doesn't exist. An option is to set it before your if/elif blocks.

---

Additionally, you can make this much shorter:

kik = "no"
age = input("age?\n")
if int(age) < 16:
    print ("Too young bby")
else:
    kik = input("Do you have kik?\n")
yes = "yes"
if kik == yes:
    print ("add me bby")
else:
    print ("bye")

Answer 2

There are a couple of things you should fix here. FIrst, store the age as a number using int():

age = int(input("age?\n"))

And then do a less than:

if(age < 16):
  print ("Too young bby")
else:
  kik = input("Do you have kik?\n")
  if kik == "yes":
    print ("add me bby")
  else:
    print ("bye")

Answer 3

Set kik to the default it should be outside your chain.

age = input("age?\n")
kik = "no" #assuming no is default
...

As it is in your code, it will only be defined if you hit else

Answer 4

The short answer is that kik is out of scope, and putting kik = "no" at the beginning of your program should get rid of that error.

But, here's a better way to do the whole thing:

age = int(input("age?\n"))
kik = "no"

if age < 16:
    print ("Too young bby")
else:
    kik = input("Do you have kik?\n")

if kik == yes:
    print ("add me bby")
else:
    print ("bye")

Answer 5

Andy's post was very helpful to me. There is one thing I would add. If you enter an integer less than 16 to the first question, you get both print ("Too young bby") and print ("bye"). The first output only is sufficient for my use, not print ("bye"). To achieve this you indent the second 'if' statement to the first 'else' statement. If you do not indent my code, you get "You are not eligible to vote in Ireland." twice. Here is my example;

Citizen = "No"
Yes = "Yes"
Age = int(input("What is your age?: \n"))
if (Age) < 18:
    print("You are not eligible to vote in Ireland.\n")
else:
    Citizen = input("Do you hold Irish Citizenship? Yes/No: \n")
    if Citizen == Yes:
        print("You are eligible to vote in Ireland.\n")
    else:
        print("You are not eligible to vote in Ireland.\n")