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 ().

Using enumerate is the standard way to go. Although, you can take advantage of the speed of str.find for time-critical operations.

Code

def find_all(s, c):
    idx = s.find(c)
    while idx != -1:
        yield idx
        idx = s.find(c, idx + 1)

print(*find_all('Apples are totally awesome', 'o')) # 12 23

I made the above return a generator for elegance and to account for very large strings. Put it can of course be casted to a list if need be.

Benchmark

Here is a benchmark against a solution using enumerate and a list-comprehension. Both solutions have linear time-complexity, but str.find is significantly faster.

import timeit

def find_all_enumerate(s, c):
    return [i for i, x in enumerate(s) if c == x]

print(
    'find_all:',
    timeit.timeit("list(find_all('Apples are totally awesome', 'o'))",
                  setup="from __main__ import find_all")
)

print(
    'find_all_enumerate:',
    timeit.timeit("find_all_enumerate('Apples are totally awesome', 'o')",
                  setup="from __main__ import find_all_enumerate")
)

Output

find_all: 1.1554179692960915
find_all_enumerate: 1.9171753468076869

The function:

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


findOccurrences(yourString, '|')

will return a list of the indices of yourString in which the | occur.

There is no simple built-in string function that does what you're looking for, but you could use the more powerful regular expressions:

import re
[m.start() for m in re.finditer('test', 'test test test test')]
#[0, 5, 10, 15]

If you want to find overlapping matches, lookahead will do that:

[m.start() for m in re.finditer('(?=tt)', 'ttt')]
#[0, 1]

If you want a reverse find-all without overlaps, you can combine positive and negative lookahead into an expression like this:

search = 'tt'
[m.start() for m in re.finditer('(?=%s)(?!.{1,%d}%s)' % (search, len(search)-1, search), 'ttt')]
#[1]

re.finditer returns a generator, so you could change the [] in the above to () to get a generator instead of a list which will be more efficient if you're only iterating through the results once.