bash scripting : printing variable with the number within $i

amdGTintel · 01-18-2007, 12:01 AM

hello.
I got this problem :
I want to print variable $1 to $n but for is not good enough in its current way, atleast how I use it :

for i in $(seq -w 1 9); do echo $$i; done
$i should give a number, but $$ is being done before so it's not good. I want it firstly to check what $i is and then do $ on the $i, if i=1 then $ $i = $1

How can I solve that?
\

Thanks!

jschiwal · 01-18-2007, 01:27 AM

Use indirection.

Code:

#!/bin/bash
for ((n=1;n<5;n++)); do
echo "${!n}"
done

./test.sh 12 43 56 78 90
12
43
56
78

Shift is usually used to handle one argument at a time.

matthewg42 · 01-18-2007, 01:30 AM

I find this is nice and easy:

Code:

#!/bin/bash

for arg in "$@"; do
    echo "we have an arg: $arg"
done

Use "$@" instead of $* to prevent problems with args which have whitespace in them.