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Python Python Collections (2016, retired 2019) Lists Shopping List Take Three

Why does my program break when I add a 'QUIT' option to exit application?

When new_item == 'DONE', my program works just fine, but why does it completely break when I change that to new_item == 'DONE' or 'QUIT'?

The problem happens on line 41: if new_item.upper() == 'DONE' or 'QUIT':

import os
shopping_list = []

def clear_screen():
    os.system("cls" if os.name =="nt" else "clear")

def num_items(shopping_list):
    print ("List has {} items.".format(len(shopping_list)))

def add_to_list(item):
    item = item.upper()
    shopping_list.append(item)
    print ("Added!")

    show_list(shopping_list)

def remove_item(item):
    item = item.upper()
    shopping_list.remove(item)

def show_help():
    clear_screen()
    print("What should we get at the store?")
    print("""
Enter 'DONE' to stop adding items.
Enter 'HELP' to show this menu.
Enter 'SHOW' to show list.
Enter 'REMOVE' to remove items
""") 

def show_list(shopping_list):
    print ("List has {} items.".format(len(shopping_list)))
    for item in shopping_list:
        print("* " + item)

#THIS IS THE LOGIC 
show_help()
while True:
    new_item = input("> ")
    if new_item.upper() == 'DONE' or 'QUIT':
        show_list(shopping_list)
        break
    elif new_item.upper() == 'HELP':
        show_help()
        continue
    elif new_item.upper() == 'SHOW':
        show_list(shopping_list)
        continue
    elif new_item.upper() == 'REMOVE':
        item = input("Which item would you like to remove?  ")
        remove_item(item)
        show_list(shopping_list)
    else:
        add_to_list(new_item)       

    ```

I think the correct syntax is

if new_item.upper() == 'DONE' or new_item.upper() == 'QUIT':

2 Answers

Steven Parker
Steven Parker
229,644 Points

As Jasper's example shows, you need to use complete expressions on both sides of the logic "or" operation. A non-empty string on one side by itself will be considered "truthy".

A more compact way of testing for multiple answers would use the membership operator ("in") with a list:

    if new_item.upper() in ['DONE', 'QUIT']:

Ok that makes sense now. It wasn't giving an error so I couldn't follow the logic very easily (still pretty new). Are you able to explain why "if new_item.upper() == 'DONE' or 'QUIT':" doesn't work logically?

The line you give Steven looks simple yet effective and I will for sure be using the format in future code.

Steven Parker
Steven Parker
229,644 Points

So a plain string (no comparison) is "truthy", so you can replace it with "True". Then anything "or"-ed with True will still be True. So:

if  new_item.upper() == 'DONE'  or  'QUIT':   # this is the same thing as....
if (new_item.upper() == 'DONE') or ('QUIT'):  # which is the same as...
if  new_item.upper() == 'DONE'  or  True:     # which is the same as...
if  True:

Does that explain it? Happy coding!

Thank you Jasper as well, I don't know how my eyes didn't read the "or new_item.upper() == 'QUIT':"

That's what you get when you are new to coding and try to type while listening.

@Payton. It happens to us all, we are ever learning. Glad that you now have a deeper understanding of what you have asked.