Counting numbers of "a" in a word.

def countA(word):
for i in range(0,len(word)):
b=0
if ord(word[i])==97:
b=b+1
return b
else:
return 0

this is my code used to count the numbers of "a" in a word. it only count small letter a and not the capital letter A. the problem is that my code cannot count words that have more a than 1 like banana whereas apple worked fine.
for example the word banana should output 3 instead my code showed only 0.Please help me thanks.

Well, take the return statements out of the loop! Now you return as soon as the first instance of “a” is found.

what do u mean taking out the return. If i took the return out the function is still not working.

I mean your function now returns as soon as it first executes “b = b + 1”. Don’t return there. Move the return statement to the end of the function.

Also, need to take the b=0 out of the for loop.

tks alot

def countA(word):
count = 0
for count, letter in enumerate(word):
if count.lower() == "a":
count += 1
return count

@ eGrove: Why do you bother to dredge up these old threads that have already been answered? The original posters are long gone and if other people don't notice the dates they may waste their time posting useless answers just like you are.

I've seen you do this over at Dream.in.Code too. Why not answer questions that NEED answering instead???