Getting the last argument passed to a shell script

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$1 is the first argument.
$@ is all of them.

How can I find the last argument passed to a shell script?

29 Answers

The simplest answer for bash 3.0 or greater is

_last=${!#}       # *indirect reference* to the $# variable
# or
_last=$BASH_ARGV  # official built-in (but takes more typing :)

That's it.

$ cat lastarg
#!/bin/bash
# echo the last arg given:
_last=${!#}
echo $_last
_last=$BASH_ARGV
echo $_last
for x; do
   echo $x
done

Output is:

$ lastarg 1 2 3 4 "5 6 7"
5 6 7
5 6 7
1
2
3
4
5 6 7

From oldest to newer solutions:

The most portable solution, even older sh (works with spaces and glob characters) (no loop, faster):

eval printf "'%s\n'" "\"\${$#}\""

Since version 2.01 of bash

$ set -- The quick brown fox jumps over the lazy dog

$ printf '%s\n'     "${!#}     ${@:(-1)} ${@: -1} ${@:~0} ${!#}"
dog     dog dog dog dog

For ksh, zsh and bash:

$ printf '%s\n' "${@: -1}    ${@:~0}"     # the space beetwen `:`
                                          # and `-1` is a must.
dog   dog

And for "next to last":

$ printf '%s\n' "${@:~1:1}"
lazy

Using printf to workaround any issues with arguments that start with a dash (like -n).

For all shells and for older sh (works with spaces and glob characters) is:

$ set -- The quick brown fox jumps over the lazy dog "the * last argument"

$ eval printf "'%s\n'" "\"\${$#}\""
The last * argument

Or, if you want to set a last var:

$ eval last=\${$#}; printf '%s\n' "$last"
The last * argument

And for "next to last":

$ eval printf "'%s\n'" "\"\${$(($#-1))}\""
dog

For bash, this comment suggested the very elegant:

echo "${@:$#}"

To silence shellcheck, use:

echo ${*:$#}

As a bonus, both also work in zsh.

To return the last argument of the most recently used command use the special parameter:

$_

In this instance it will work if it is used within the script before another command has been invoked.

$ echo "${*: -1}"

That will print the last argument

With GNU bash version >= 3.0:

num=$#                 # get number of arguments
echo "${!num}"         # print last argument

Just use !$.

$ mkdir folder
$ cd !$ # will run: cd folder
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