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# asks the question and gives the choices   
choice=input('what is your favourite colour? press 1 for red, 2 for blue, 3 for yellow or 4 to quit') 
# if the responder responds with choice 1 or 2,3,4 it will print this
if choice=='1' or 'one':
    print('red means anger') 

elif choice=='2'or 'two':
    print('blue represents calmness')
elif choice=='3' or 'three':
    print('yellow represents happiness')
elif choice=='4' or 'four':
    print('it was nice meeting you. goodbye.')
else:
    print('sorry incorrect answer please try again.')

One of my students wrote this and I can't seem to get it working. HELP! It keeps repeating red means anger. If I comment out the 'or', it works but why can't she use 'or'? I want her to add a loop but only if this works first.

Skycc
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Lisa G
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2 Answers2

3

The or is not used correctly. You need to write 

choice == '1' or choice == 'one'

Otherwise type coercion will evaluate 'one' to true and the or condition of the first if statement is always true (a tautology) and the other cases are never checked.

DivineTraube
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when you have statement like 

if choice=='1' or 'one':

it will be treated as 

if (choice=='1') or 'one':

'one' is always evaluate to True, thus the if condition is always fulfill, what you need is as below, bracket to make things clearer

if (choice=='1') or (choice=='one'):

Alternatively, when you have multiple OR statement to be check against a value, can consider put all the check value in a list and use in like below, this look cleaner to me when the value to check increasing

if choice in ['1', 'one', '2', 'two', '3', 'three']:
Skycc
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  • @David Schwartz, thanks for pointing that out, added some explanation to original question though its already answer by DivingTraube, suggest the later as alternative – Skycc Nov 09 '16 at 00:36