Getting the last argument passed to a shell script – Dev

The best answers to the question “Getting the last argument passed to a shell script” in the category Dev.

QUESTION:

$1 is the first argument.
[email protected] is all of them.

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

ANSWER:

This is a bit of a hack:

for last; do true; done
echo $last

This one is also pretty portable (again, should work with bash, ksh and sh) and it doesn’t shift the arguments, which could be nice.

It uses the fact that for implicitly loops over the arguments if you don’t tell it what to loop over, and the fact that for loop variables aren’t scoped: they keep the last value they were set to.

ANSWER:

This is Bash-only:

echo "${@: -1}"

ANSWER:

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

ANSWER:

$ set quick brown fox jumps

$ echo ${*: -1:1} # last argument
jumps

$ echo ${*: -1} # or simply
jumps

$ echo ${*: -2:1} # next to last
fox

The space is necessary so that it doesn’t get interpreted as a default value.

Note that this is bash-only.