How to find char in string and get all the indexes?

This is because str.index(ch) will return the index where ch occurs the first time. Try:

def find(s, ch):
    return [i for i, ltr in enumerate(s) if ltr == ch]

This will return a list of all indexes you need.

P.S. Hugh's answer shows a generator function (it makes a difference if the list of indexes can get large). This function can also be adjusted by changing [] to ().

I would go with Lev, but it's worth pointing out that if you end up with more complex searches that using re.finditer may be worth bearing in mind (but re's often cause more trouble than worth - but sometimes handy to know)

test = "ooottat"
[ (i.start(), i.end()) for i in re.finditer('o', test)]
# [(0, 1), (1, 2), (2, 3)]

[ (i.start(), i.end()) for i in re.finditer('o+', test)]
# [(0, 3)]

Lev's answer is the one I'd use, however here's something based on your original code:

def find(str, ch):
    for i, ltr in enumerate(str):
        if ltr == ch:
            yield i

>>> list(find("ooottat", "o"))
[0, 1, 2]

def find_offsets(haystack, needle):
    """
    Find the start of all (possibly-overlapping) instances of needle in haystack
    """
    offs = -1
    while True:
        offs = haystack.find(needle, offs+1)
        if offs == -1:
            break
        else:
            yield offs

for offs in find_offsets("ooottat", "o"):
    print offs

results in

0
1
2

def find_idx(str, ch):
    yield [i for i, c in enumerate(str) if c == ch]

for idx in find_idx('babak karchini is a beginner in python ', 'i'):
    print(idx)

output:

[11, 13, 15, 23, 29]