Get last parameter on shell script [duplicate]

Answer 1

For this, you can use:

${@: -1}

Test

$ cat a
#!/bin/bash

echo "passed $# parameters, last being --> ${@: -1}"

$ ./a a b c d
passed 4 parameters, last being --> d
$ ./a a b c d e f g
passed 7 parameters, last being --> g

Answer 2

It might be better to reorganize your parameters. Instead of a variable number of arguments followed by a specific final argument, put the last argument first (so that it is $1), and put the variable number of arguments after that. So instead of

myshell_script a b c d

with var2 eventually being set to "d", use

var2=$1; shift

and call like

myshell_script d a b c

Now var2 will have the value the value of "d", and $@ (after the shift) will contain "a", "b", and "c".

Answer 3

The script below shows how you can get the first and last arguments passed to a script:

numArgs="$#"
echo "Number of args: $numArgs"

firstArg="$1"
echo "First arg: $firstArg"

lastArg="${!#}"
echo "Last arg: $lastArg"

Output:

$ ./myshell_script.sh a b c d e f
Number of args: 6
First arg: a
Last arg: f

Answer 4

Quoting a way from here:

for last; do : ; done
echo "${last}"

The last argument passed to the script would be stored in the variable last.

As mentioned in the link, this would work in POSIX-compatible shells it works for ANY number of arguments.

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BTW, I doubt if your script works the way you've written in your question:

var1 = `echo "$#"`

You need to remove those spaces around =, i.e. say:

var1=`echo "$#"`

or

var1=$(echo "$#")