BASH SCRIPT : How to use a variable inside another variable

Question

I need to store command line arguments passed in an array

my Command is

./test1.sh 2 4 6

Now i need to store 2 4 6 in an array and im using..

s1=$#     
"it tells how many arguments are passed."


for (( c=1; c<=$s1; c++ ))

do

     a[$c]}=${$c}

I have written ${$c} for taking the arguments value but it is showing bad substition.

johangu · Accepted Answer · 2011-08-11 11:39:13Z

7

This will give you an array with the arguments: args=("$@")

And you can call them like this: echo ${args[0]} ${args[1]} ${args[2]}

answered Aug 11, 2011 at 11:39

johangu

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1 Comment

Glad I could help. Could you mark an accepted answer, so the issue wont stay in the "unanswered" section?

Zskdan · Accepted Answer · 2011-08-11 11:59:22Z

1

bash variable can be indirectly referenced by \$$VARNAME or by ${!VARNAME} in version 2 so your assignment statement must be :

a[$c]=${!c}

or

eval a[$c]=\$$c

answered Aug 11, 2011 at 11:59

Zskdan

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Karoly Horvath · Accepted Answer · 2011-08-11 11:44:00Z

0

You can always pick the first argument and drop it. Use $1 and shift.

c=1
while [ $# -gt 0 ]; do
    a[$c]=$1
    c=$((c+1))
    shift
done

answered Aug 11, 2011 at 11:38

Karoly Horvath

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