While Loop Conditional statement only calls the first condition

Question

I am experiencing unwanted behavior with my code's if statements. When I call conditional statements inside the while loop, only the first condition is called no matter which input is supplied. I wish for the proper input to call the properly described function and then continue back to the beginning of the loop to ask the user for another choice.

color_modes = ['sangria', 'ham', 'nightHawk']
print color_modes
def sangria():
    my_range = list(range(20))
    print my_range
def ham():
    print 'foo'
def nightHawk():
    print 'nightHawk'
while True:
    user_input = input('...')
    if 'sangria':
        ham()
        continue
    if 'ham':
        sangria()
        continue
    if 'nightHawk':
        nightHawk()
        continue
    else:
        break

Moreoever, when I use the syntax:

if user_input == 'ham': instead of the shorthand if 'ham:, the condition does not function. Thank you in advance.

mhawke · Accepted Answer · 2016-12-03 10:56:02Z

In Python 2

user_input = input('...')

attempts to evaluate the user's input. If you input a string such as ham Python will try to evaluate the expression ham which will fail as shown below (unless there is a similarly named variable in scope).

>>> input('...')
...ham
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<string>", line 1, in <module>
NameError: name 'ham' is not defined

You could fix it by using raw_input() instead:

user_input = raw_input('...')

and then using string comparison:

while True:
    user_input = raw_input('...')
    if user_input == 'ham':
        ham()
    elif user_input == 'sangria':
        sangria()
    elif user_input == 'nightHawk':
        nightHawk()
    else:
        break

This code also uses elif to avoid the unnecessary continue statements.

A better way is to use a dictionary to map the user input to the associated function:

input_map = {'ham': ham, 'sangria': sangria, 'nightHawk': nightHawk}
while True:
    func = input_map.get(raw_input('...'))
    if func is None:
        break
    func()

OneCricketeer · Accepted Answer · 2016-12-03 10:46:43Z

1

if 'sangria': is only shorthand for if bool('sangria'):, which always evaluates to if True, therefore the first thing is ran regardless of input

You need to compare the input string. There's no shorthand / alternative way to write it like a Java switch case statement

This code looks odd, by the way.

if 'sangria':
    ham()
    continue
if 'ham':
    sangria()
    continue

If I understand what you want, you can create a function map.

funcs = {"ham": ham, "sangria" : sangria, "nightHawk" : nightHawk} 
user_input = input() 
funcs[user_input]() # use the string to get the function object, then call it

edited Dec 3, 2016 at 10:46

answered Dec 3, 2016 at 10:39

OneCricketeer

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Comments

User2342342351 · Accepted Answer · 2016-12-03 18:40:41Z

1

if 'ham' will always pass since an not empty string is always True.

Instead, you must check for equality using the == operator.

The correct code would be:

color_modes = ['sangria', 'ham', 'nightHawk']
print color_modes
def sangria():
    my_range = list(range(20))
    print my_range
def ham():
    print 'foo'
def nightHawk():
    print 'nightHawk'
while True:
    user_input = input('...')
    if user_input == 'sangria':
        ham()
        continue
    if user_input == 'ham':
        sangria()
        continue
    if user_input == 'nightHawk':
        nightHawk()
        continue
    else:
        break

edited Dec 3, 2016 at 18:40

answered Dec 3, 2016 at 10:39

User2342342351

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1 Comment

Erik Sven Broberg Over a year ago

Thank you for the reply. i understand now that raw_input in 2.7 and input in python 3 return strings, but input by default in 2.7 returns an expression. I suppose it was this simple error on my part. Thank you to all for clarifiying. Thank you as well for clarifying what if 'somestring' does.

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