Bash indexof substring after nth character

Question

is there a way to get indexOf of a substring after nth character in Bash, without needing to cut the original string one by one, that would be lengthy

              12345678901234
$ expr index 'abcdeabcdabcaa' 'c'
3
$ expr index 'abcdeabcdabcaa' 'ca'
1

What i want would be:

$ indexOf 'abcdeabcdabcaa' 'ca'
12
$ indexOfAfter 5 'abcdeabcdabcaa' 'c'
8
$ indexOfAfter 5 'abcdeabcdabcaa' 'ca'
12
$ indexOfAfter 9 'abcdeabcdabcaa' 'b'
11
$ indexOfAfter 111 'abcdeabcdabcaa' 'c'
0

probably there are already function on Bash to do this.. this is not a homework, just out of curiosity..

possible duplicate of [Bash: Find position of character in a string under OS X](http://stackoverflow.com/questions/17615750/bash-find-position-of-character-in-a-string-under-os-x) — chepner, Jul 17 '13 at 12:00

suspectus · Accepted Answer · 2013-07-17T09:18:40.240

6

both operations implemented with awk

contents of file indexOf:

echo "$1" "$2" | awk '{print index($1,$2)}'

e.g.

 indexOf 'abcdeabcdabcaa' 'ca'
 12

contents of file indexOfAfter:

echo "$1" "$2" "$3" | \
awk '{s=substr($2,$1);posn=index(s,$3);if (posn>0) print $1+posn-1; else print 0;}'

edited Jul 17 '13 at 09:18

answered Jul 17 '13 at 09:09

suspectus

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Bash indexof substring after nth character

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