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Start your free trialPayton Roedel
1,555 PointsWhy 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)
```
2 Answers
Steven Parker
231,275 PointsAs 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']:
Payton Roedel
1,555 PointsOk 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
231,275 PointsSo 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!
Payton Roedel
1,555 PointsThank 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.
Jasper Maposa
26,918 Points@Payton. It happens to us all, we are ever learning. Glad that you now have a deeper understanding of what you have asked.
Jasper Maposa
26,918 PointsJasper Maposa
26,918 PointsI think the correct syntax is
if new_item.upper() == 'DONE' or new_item.upper() == 'QUIT':