def loopy(items): if == 'a' print(items) def loopy(items): if != 'a' print(items)

1. The code inside a function needs to start on the next line after the colon and needs to be indented:

def loopy(items):
    # code starts here

1. You'll need a for loop to iterate through each item in items
2. Inside the loop, check each item to see if it begins with 'a'. You'll need an if/else for this so the loop knows whether to continue to the next item, or print it. To see if the item begins with 'a' check the character at index 0. You can get the index of something with square brackets:

first_letter = some_word[0]

In the end your logic should look something like this:

def loopy(items):
    # for each item in items:
        #if the first letter of the current item is a:
            #continue
        #else:
            # print the item

Thanks a ton ...that was expertly put and a great answer .!!! I asked this question months ago.. bottomline you deserve good credit for your answer. Thanks again. Henry

Thanks Henry, glad to help! Consider hitting the old up-vote button if you feel it was helpful. Happy coding and good luck!

what's wrong with my code

I get Didn't find the right items being printed

def loopy(items):
    for word in items:
        if items[0] == "a":
            continue
        else:
             print(items)

You are iterating through with the items rather than the word.

if word[0] == 'a':