create variable name that contains a variable 1

[pavNell]

trying to create new variables whose names contain a variable. For instance, I've tried...


x=2
while test $x -ge 1
do
var_$x="Hello $x"
x=`expr $x - 1`
done

echo $var_2
echo $var_1


this script outputs...

var_2=Hello 2: not found
var_1=Hello 1: not found

thanks for any help.

 

[feherke]

Hi

In your shell man page ( you forget to mention its name... ) look for the [tt]eval[/tt].

Feherke.

http://rootshell.be/~feherke/

 

[pavNell]

I'm using the bash shell on my zaurus.

I'm not getting it. Guess it's the syntax for eval that I don't understand. I've read over the man page and tried several different ways to use eval. Help please???

 

[feherke]

Hi

Example. The [tt]bash[/tt] internal variables for the command prompts are [tt]PS1[/tt], [tt]PS2[/tt], [tt]PS3[/tt] and [tt]PS4[/tt]. You want to display their values in a loop, then assign them new values as (1)>, (2)>, (3)> and (4)> respectively.

for ((i=1;i<=4;i++)); do eval "echo PS$i = \$PS$i"; done

for ((i=1;i<=4;i++)); do eval "PS$i=\"($i)> \""; done

Feherke.

http://rootshell.be/~feherke/

 

[pavNell]

Thank you feherke, it worked!

My attempts were close. I didn't add or escape the extra double-quotes. Also the for loop looks cleaner than using a while loop.

A star for your help!