How to get the last command line parameter?


 
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# 1  
Old 02-18-2010
How to get the last command line parameter?

"$#" gives the number of command-line arguments. How do you get the last command-line parameter (or any particular one determined by a variable)? I thought it would be "${$#}", but that produces something completely unexpected.
# 2  
Old 02-18-2010
Not sure it's the best way, but...
Code:
$ cat Test
eval echo \$$#

$ ./Test a b c
c

To get the one-from-last...
Code:
eval echo \$$(($#-1))

Otherwise, simply $1, $2, $3, etc...

The completely unexpected number you mention is the process id of the "current" process.

Last edited by Scott; 02-18-2010 at 07:49 PM..
# 3  
Old 02-18-2010
I typically set a variable to get the last argument
Code:
let last=$#-1
echo "$last"

# 4  
Old 02-19-2010
Solution

Both will not give the exact answer he want .Here is the solution .

eval echo \$$(($#))
# 5  
Old 02-21-2010
If you have more than 9 arguments, you need { }.
Code:
eval echo \${$#}

Or looping all args
Code:
while [ $# -gt 0 ]
do
      last="$1"
      shift
done

Or put args to the array
Code:
args=( "$@" )
cnt="${#args[@]}"
# 1st index 0
(( lastfld=cnt-1 ))
last="${args[$lastfld]}"
echo $last

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