Assign last character of string to variable in bash script [duplicate] - bash

Question

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?

Answer 1

Last three characters of string:
${string: -3}
or
${string:(-3)}
(mind the space between : and -3 in the first form).
Please refer to the Shell Parameter Expansion in the reference manual:
${parameter:offset}
${parameter:offset:length}
Expands to up to length characters of parameter starting at the character
specified by offset. If length is omitted, expands to the substring of parameter
starting at the character specified by offset. length and offset are arithmetic
expressions (see Shell Arithmetic). This is referred to as Substring Expansion.
If offset evaluates to a number less than zero, the value is used as an offset
from the end of the value of parameter. If length evaluates to a number less than
zero, and parameter is not ‘#’ and not an indexed or associative array, it is
interpreted as an offset from the end of the value of parameter rather than a
number of characters, and the expansion is the characters between the two
offsets. If parameter is ‘#’, the result is length positional parameters
beginning at offset. If parameter is an indexed array name subscripted by ‘#’ or
‘*’, the result is the length members of the array beginning with
${parameter[offset]}. A negative offset is taken relative to one greater than the
maximum index of the specified array. Substring expansion applied to an
associative array produces undefined results.
Note that a negative offset must be separated from the colon by at least one
space to avoid being confused with the ‘:-’ expansion. Substring indexing is
zero-based unless the positional parameters are used, in which case the indexing
starts at 1 by default. If offset is 0, and the positional parameters are used,
$# is prefixed to the list.
Since this answer gets a few regular views, let me add a possibility to address John Rix's comment; as he mentions, if your string has length less than 3, ${string: -3} expands to the empty string. If, in this case, you want the expansion of string, you may use:
${string:${#string}<3?0:-3}
This uses the ?: ternary if operator, that may be used in Shell Arithmetic; since as documented, the offset is an arithmetic expression, this is valid.
Update for a POSIX-compliant solution
The previous part gives the best option when using Bash. If you want to target POSIX shells, here's an option (that doesn't use pipes or external tools like cut):
# New variable with 3 last characters removed
prefix=${string%???}
# The new string is obtained by removing the prefix a from string
newstring=${string#"$prefix"}
One of the main things to observe here is the use of quoting for prefix inside the parameter expansion. This is mentioned in the POSIX ref (at the end of the section):
The following four varieties of parameter expansion provide for substring processing. In each case, pattern matching notation (see Pattern Matching Notation), rather than regular expression notation, shall be used to evaluate the patterns. If parameter is '#', '*', or '#', the result of the expansion is unspecified. If parameter is unset and set -u is in effect, the expansion shall fail. Enclosing the full parameter expansion string in double-quotes shall not cause the following four varieties of pattern characters to be quoted, whereas quoting characters within the braces shall have this effect. In each variety, if word is omitted, the empty pattern shall be used.
This is important if your string contains special characters. E.g. (in dash),
$ string="hello*ext"
$ prefix=${string%???}
$ # Without quotes (WRONG)
$ echo "${string#$prefix}"
*ext
$ # With quotes (CORRECT)
$ echo "${string#"$prefix"}"
ext
Of course, this is usable only when then number of characters is known in advance, as you have to hardcode the number of ? in the parameter expansion; but when it's the case, it's a good portable solution.

Answer 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

Answer 3

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.

Answer 4

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, ${#string} is the length of the string, and the trailing "-" means cut to the end.
3. Alternative using awk
This solution instead uses the substring function of awk to select a substring which has the syntax substr(string, start, length) going to the end if the length is omitted. length($string)-2) thus picks up the last three characters.
echo $string | awk '{print substr($1,length($1)-2) }'