how to make bash replace the value of a variable?

rabbit2345 · 04-18-2009, 09:49 PM

Hi,

I have a program I am writing where I have a for loop and I want to make it substitute the variable twice like:

for ((i=0;i<5;i++)) do
echo $"$i"
done

I wanted it to replace $i with a value, and then echo the value of the variable named from that output. How can you tell bash to do a double substitution?

Like:
echo $"$i"

echo $3

echo [text from $3]

Thanks,
rabbit2345

rriggs · 04-18-2009, 10:26 PM

http://www.enricozini.org/2008/tips/...direction.html

bartonski · 04-18-2009, 10:33 PM

I'm not sure how to do exactly what you're asking, but I think that you could get the same effect using an array

Code:

#! /bin/bash
declare MY_ARRAY=("foo" "bar" "baz" "quux" "quuux")

for ((i=0;i<5;i++)) do
echo ${MY_ARRAY[$i]}
done

rabbit2345 · 04-18-2009, 11:02 PM

thanks rriggs for your reply. It worked out exactly the way I needed it.

Thanks,
rabbit2345