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The shell command and any arguments to that command appear as numbered shell variables: $0 has the string value of the command itself, something like script , ./script , /home/user/bin/script or whatever. Any arguments appear as "$1" , "$2" , "$3" and so on. The count of arguments is in the shell variable "$#" .
$# : This variable contains the number of arguments supplied to the script. $? : The exit status of the last command executed. Most commands return 0 if they were successful and 1 if they were unsuccessful.
In the command-line, you can press alt + . or esc - . It cycles through the last argument of your previous commands.
"$@" Stores all the arguments that were entered on the command line, individually quoted ("$1" "$2" ...). So basically, $# is a number of arguments given when your script was executed. $* is a string containing all arguments. For example, $1 is the first argument and so on.
This is Bash-only:
echo "${@: -1}"
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.
$ set quick brown fox jumps
$ echo ${*: -1:1} # last argument
jumps
$ echo ${*: -1} # or simply
jumps
$ echo ${*: -2:1} # next to last
fox
The space is necessary so that it doesn't get interpreted as a default value.
Note that this is bash-only.
The simplest answer for bash 3.0 or greater is
_last=${!#} # *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
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