Access Bash Positional Parameter Through Variable

accessing a positional parameter through a variable

Do

[[ "$var" -eq "1" ]] && echo "$1" #for positional parameter 1

and so on or you could do it like below :

#!/bin/sh
var=2
eval echo "\$${var}" # This will display the second positional parameter an so on

Edit

how do I use this if I actually need to access outside of echo. for
example "if [ -f \$${var} ]"

The method you've pointed out it the best way:

var=2 
eval temp="\$${var}"
if [ -f "$temp" ] # Do double quote
then
#do something
fi

Access positional parameters from within a function in Bash

Just for the record, this is possible in extdebug mode through BASH_ARGC and BASH_ARGV variables, but populating a global array with positional parameters is much easier than that, and has no side-effects.

$ bash -O extdebug -s foo bar
$ f() {
>   declare -p BASH_ARGC BASH_ARGV
>   echo ${BASH_ARGV[BASH_ARGC[0] + BASH_ARGC[1] - 1]}
> }
$ f 42 69
declare -a BASH_ARGC=([0]="2" [1]="2")
declare -a BASH_ARGV=([0]="69" [1]="42" [2]="bar" [3]="foo")
foo

assign last positional parameter to variable and remove it from $@

Use a loop to move all arguments except the last to the end of positional parameters list; so that the last becomes the first, could be referred to by $1, and could be removed from the list using shift.

#!/bin/sh -
count=0
until test $((count+=1)) -ge $#
do
  set -- "$@" "$1"
  shift
done

LAST=${1-}
shift $((!!$#))

echo "$LAST"
echo "$@"

Positional parameter not readable inside the variable - bash script

If you invoke your script as ./mycript.sh $michael and the variable michael is not set in the shell, they you are calling your script with no arguments. Perhaps you meant ./myscript.h michael to pass the literal string michael as the first argument. A good way to protect against this sort of error in your script is to write:

#!/bin/bash
var1=$( linux command to list ldap users | grep "user: ${1:?}")
echo "$var1"

The ${1:?} will expand to $1 if that parameter is non-empty. If it is empty, you'll get an error message.

If you'd like the script to terminate if no values are found by grep, you might want:

var1=$( linux command to list ldap users | grep "user: ${1:?}") || exit

But it's probably easier/better to actually validate the arguments and print an error message. (Personally, I find the error message from ${:?} constructs to bit less than ideal.) Something like:

#!/bin/bash
if test $# -lt 1; then echo 'Missing arguments' >&2; exit 1; fi

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