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haxor

Member

Registered: 2010-04-19

Posts: 1

Bash var within a varname

Hi im having this problem with some varnames..
the thing is: i've set a bunch of variables from a file splitting it into a new var at each newline. it uses a couner at the end
the code i use for setting the var is as following

exec 3< filename
   while read <&3
      do echo "$REPLY" > /dev/NULL
      export VAR$COUNTER="$REPLY"
      COUNTER=`echo $COUNTER + 1 | bc`
   done
exec 3>&-

The problem i'm having is that when (a bunch of code later) i need these vars the seem not so usefull...

$COUNTER=3
for i in `seq 1 $COUNTER`;
  do
    echo "saving output"
    echo "$VAR$i"
  done

i thought this code would result into the following

echo $VAR1
echo $VAR2
echo $VAR3

instead all the code outputs is

1
2
3

I know it's looking for $VAR who doesn't exist
but how can i fix it, what am i doing wrong??

Thanks in advance

Busata

Member

From: Belgium

Registered: 2010-04-04

Posts: 30

Website

Re: Bash var within a varname

are you sure you're not looking for arrays? http://tldp.org/LDP/abs/html/arrays.html

davvil

Member

Registered: 2008-05-06

Posts: 165

Re: Bash var within a varname

You can use eval. For setting the variables:

eval export VAR$COUNTER="$REPLY"

(as a side note, do you need to export the variable?)

And for getting the value:

eval echo \$VAR$i

Or yes, use arrays, as Busata suggests.

Last edited by davvil (2010-04-19 19:13:40)

delerious010

Member

From: Montreal

Registered: 2008-10-07

Posts: 72

Re: Bash var within a varname

set a dynamically named variable : eval varname$i=$value
retrieve a dynamically named variable : ${!varname$i}

Profjim

Member

From: NYC

Registered: 2008-03-24

Posts: 658

Re: Bash var within a varname

Also you want /dev/null, not /dev/NULL