[string] Accessing last x characters of a string in Bash

I found out that with ${string:0:3} one can access the first 3 characters of a string. Is there a equivalently easy method to access the last three characters?

Another workaround is to use grep -o with a little regex magic to get three chars followed by the end of line:

$ foo=1234567890
$ echo $foo | grep -o ...$
890

To make it optionally get the 1 to 3 last chars, in case of strings with less than 3 chars, you can use egrep with this regex:

$ echo a | egrep -o '.{1,3}$'
a
$ echo ab | egrep -o '.{1,3}$'
ab
$ echo abc | egrep -o '.{1,3}$'
abc
$ echo abcd | egrep -o '.{1,3}$'
bcd

You can also use different ranges, such as 5,10 to get the last five to ten chars.

1. Generalized Substring

To generalise the question and the answer of gniourf_gniourf (as this is what I was searching for), if you want to cut a range of characters from, say, 7th from the end to 3rd from the end, you can use this syntax:

${string: -7:4}

Where 4 is the length of course (7-3).

2. Alternative using cut

In addition, while the solution of gniourf_gniourf is obviously the best and neatest, I just wanted to add an alternative solution using cut:

echo $string | cut -c $((${#string}-2))-

Here, ${#string} is the length of the string, and the "-" means cut to the end.

3. Alternative using awk

This solution instead uses the substring function of awk to select a substring which has the syntax substr(string, start, length) going to the end if the length is omitted. length($string)-2) thus picks up the last three characters.

echo $string | awk '{print substr($1,length($1)-2) }'

You can use tail:

$ foo="1234567890"
$ echo -n $foo | tail -c 3
890

A somewhat roundabout way to get the last three characters would be to say:

echo $foo | rev | cut -c1-3 | rev