Modifying a parameter pass to a script (Bash)

Question

I have been looking on Google for quite a while now and can\'t find anything that is matching what I need/want to do.

My objective is to write a script that takes tw

Answer 1

To set the positional parameters $1, $2, ..., use the set command:

set foo bar baz
echo "$*"   # ==> foo bar baz
echo $1     # ==> foo

set abc def
echo "$*"   # ==> abc def

If you want to modify one positional parameter without losing the others, first store them in an array:

set foo bar baz
args=( "$@" )
args[1]="BAR"
set "${args[@]}"
echo "$*"   # ==> foo BAR baz

Answer 2

adymitruk already said it, but why do you want to assign to a parameter. Woudln't this do the trick?

if `echo :$1: | grep ":$2:" 1>/dev/null 2>&1`
then
  echo $1
else
  echo $1:$2
fi

Maybe this:

list="1:2:3:4"
list=`./script $list 5`;echo $list

BIG EDIT:

Use this script (called listadd for instance):

if ! `echo :${!1}: | grep ":$2:" 1>/dev/null 2>&1`
then
  export $1=${!1}:$2
fi

And source it from your shell. Result is the following (I hope this is what wsa intended):

lorenzo@enzo:~$ list=1:2:3:4
lorenzo@enzo:~$ source listadd list 3
lorenzo@enzo:~$ echo $list
1:2:3:4
lorenzo@enzo:~$ source listadd list 5
lorenzo@enzo:~$ echo $list
1:2:3:4:5
lorenzo@enzo:~$ list2=a:b:c
lorenzo@enzo:~$ source listadd list2 a
lorenzo@enzo:~$ echo $list2
a:b:c
lorenzo@enzo:~$ source listadd list2 d
lorenzo@enzo:~$ echo $list2
a:b:c:d

Answer 3

First copy your parameters to local variables:

Arg1=$1

Then, where you are assigning leave off the $ on the variable name to the left of the =

You can't have a $ on the left hand side of an assignment. If you do, it's interpreting the contents of $1 as a command to run

hope this helps