variable substitution


 
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# 1  
Old 05-16-2003
Question variable substitution

Hi everyone,

I have a simple question to ask :
In a script that I'm writting, I need to create variables on-the-fly.
For instance, for every iterartion of the following loop a var_X variable should be generated :

#!/bin/ksh
a="1 2 3"
for i in $a
do
var_${i}=$i
echo "${var_$i}"
done

The above lines result in :
var_a=a: not found
"${var_$i}": bad substitution

My question is :
Is it OK to use a variable as part of another variable's name?
If yes how may I overcome the above?

Thank you all in advance,
CK
# 2  
Old 05-16-2003
You really should be using an array:

var[i]=$i
echo ${var[i]}

However, you can do exactly what you want:

eval var_$i=\$i
eval echo \$var_$i
 
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