If Statement doesn't read input

Question

Just learning how to code, and wanted to make a small program to see what I know.

n = int(input("Pick a number any Number: "))
if  n > 100:
    print ("No... Not that Number")
else:
    answer = input("Would you like to know your number?")
    if answer == "Y" or "Yes" or "y" or "yes":
        print ("Your number is %s" % (n))
    elif answer == "N" or "No" or "n" or "no" or "NO":
        print ("Oh, well that's a shame then.")
    else:
        print ("Please type Yes or No")

input("Press Enter/Return to Exit")

Everything works, except for the second if statement, which doesn't follow any of the data entered into input. Any reason why it does this?

correct the indentation first – Arpit Solanki Jun 12 '17 at 06:22 — Arpit Solanki, Jun 12 '17 at 06:22
Is that the whole code or is this the body of a while loop? – syntaxError Jun 12 '17 at 06:24 — syntaxError, Jun 12 '17 at 06:24

score 1 · Answer 1 · answered Jun 12 '17 at 06:28

1

Python isn't human, it doesn't understand

 if answer == "Y" or "Yes"

The way you mean it to. You should do

if answer == 'Y' or answer == 'Yes'

Or even better

if answer in ('Yes', 'Y', 'yes', 'y')

Or even shorter

if answer.lower() in ('yes', 'y')

answered Jun 12 '17 at 06:28

Ofer Sadan

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score 0 · Answer 2 · answered Jun 12 '17 at 06:25

== has a higher precedence than or. So, in the if condition your're checking whether answer == 'Y' and then oring this boolean expression with "Yes", which is a non-None string, so it evaluates as True. Instead, you should use the in operator to check if answer is one of the values you're interested in:

if answer in ("Y", "Yes", "y", "yes"):
    print ("Your number is %s" % (n))
elif answer in ("N", "No", "n", "no", "NO"):
    print ("Oh, well that's a shame then.")
else:
    print ("Please type Yes or No")

Even if the precedence was altered, it would not work; `or` simply doesn't follow English grammar rules. — Martijn Pieters, Jun 12 '17 at 06:26

If Statement doesn't read input

2 Answers2