Welcome to the Treehouse Community

Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.

Start your free trial

Python Python Basics (2015) Shopping List App Break

Posted by Yusuf Gillani
.a{fill-rule:evenodd;}techdegree
Yusuf Gillani
Python Web Development Techdegree Student 3,643 Points

This is still not working for me :(

Can someone tell me what I'm doing wrong? I've put the if statement inside the loop.

breaks.py 
 def loopy(items):
    for item in items:
        if item == "STOP":
            break

1 Answer

AJ Salmon
AJ Salmon
5,675 Points
AJ Salmon
AJ Salmon
5,675 Points

Your if statement is correct! The issue is that the function is still supposed to print any items that come before 'STOP'. Here's an example of what the output should be for two similar lists:

>>> list1 = ['apple', 3, True, 'Star Wars']
>>> loopy(list1)
apple
3
True
Star Wars
>>> list2 = ['apple', 3, 'STOP', True, 'Star Wars']
>>> loopy(list2)
apple
3

Because there was no 'STOP' string in list1, all of the items were printed. The for loop broke when it got to the end of the list. When loopy is passed list2 as the argument, it prints all of the items that come before 'STOP'. All you need to do is make sure your function is still printing the items it's given! Hope this helps