Need HELP with this function!

In this function i have 3 strings...

def update_view (puzzle, view, guessed_letter):
''' (str, str, str) -> str

Return the view of the puzzle with each occurrence of the letter in
the puzzle revealed.

>>>update_view ('apple', '^^^^^', 'a')
'a^^^^'
>>>update_view ('apple', 'a^^^^', 'p')
'app^^'

'''

My function so far is:
result = ""
for i in range(len(puzzle)):
c = puzzle[i]
if c.islower() and c not in guessed_letter:
result+= HIDDEN
else:
result = result + c
return result

But this is not returning the 'view' parameter with the updated guessed letters, instead it is assuming that the whole function is hidden and updating only the guessed letters for example:

This is what it should RETURN!

>>>update_view ('apple', 'a^^^^', 'p')
'app^^'

THIS IS WHAT IT IS RETURNING:

>>>update_view ('apple', 'a^^^^', 'p')
'^pp^^' <-- with 'a' hidden again...

PLEASE HELP!

Whoot! You're the first poster I recall (other than myself) to doctest. Congratulations. So very nice to have tests that verify what your function should do. See the link at my signature to post executable python code.

Anyway, you'll need to return the next character from the view or from the puzzle, whereas you using HIDDEN instead of view. You may have to rethink your start-up, such that the initial view is

view = HIDDEN*len(puzzle)

Is there a way to do this without using enumerate??? Because it isnt working, and i also set the variable for view as view = HIDDEN* len(puzzle) I tried it but it just wont work, its coming out as the updated version again, and not with the view + updated view with guessed letter((((

You must have a fairly old version of python if it hasn't got enumerate. The main idea was to use the next character from view rather than HIDDEN. But of course, you can just go back to your

OMG... You are a life saver!! Thank You soooo much