Fetching last n characters of string in ksh
ksh, linux, shell, solaris, string
Solution
You can use this, say n=2 (we will use 2 question marks in the nested expansion):
$ var="this is my var"
$ echo "${var#${var%??}}"
ar
Explanation
It is a nested expansion.
The expansion `${var%%??}` is embedded in the expansion `${var# }`. The `${var#string}` expansion will cut off anything from the beginning of the variable which matches 'string'. So we are saying in this instance, removing anything from the beginning of the variable which matches `${var%%??}`.
On its own, `${var%%??}` matches "this is my v" for the variable in the example, as the`%%` expansion matches the longest possible match at the end of the variable. In this case, two regexp ?'s.
Problem
Let say i have a variable `name` holding some string value To fetch last n characters, in bash we write : ``` $ echo "${name: -n}" ``` what is the equivalent way in `ksh`, i have seen `sed` and `awk` methods but what i am looking for is one line or piping solution similar to `bash` to extract last characters These are errors and efforts so far : ``` AB12 $ name="123456" AB12 $ echo ${name:(-3)} ksh: ${name:(-3)}: bad substitution AB12 $ echo${name:0:-3} ksh: echo${name:0:-3}: bad substitution AB12 $ print ${name%?} 12345 AB12 $ echo "some string" | tail -c -1 tail: cannot open input AB12 $ echo -n "my string discard"| tail -c -1 tail: cannot open input AB12 $ echo "foo"| cut -c 1-2 fo AB12 $ echo "foo"| cut -c -2 fo AB12 $ echo $name 123456 AB12 $ echo "${name: -3}" ksh: "${name: -3}": bad substitution ``` I am on Solaris currently - if this helps!