KSH variable substitution

Hi,

I am trying to substitute a variable with another variable like below:

for x in 1 2 3 4
do
sourcedir=datadir${x}
echo $sourcedir
filelist=`ls ${${sourcedir}}`
echo $filelist
sleep 60

But this doesnt work. $sourcedir = datadir1 which is correct.

datadir1 is a var pointing to a directory

but filelist=`ls ${${sourcedir}}` doesnt work

I want to ls the var $datadir1 but how do I do this given the way datadir1 ic constructed ??

Thanks in advance

hey Huck,

filelist='ls ${sourcedir}'

or better

filelist='ls $sourcedir'

-Steven

Hi Steven,

thanks for your help but that just tries to list datadir1 (e.g. ls datadir1). What I want it to do is ls $datadir1 which is a var defined and which points to a dir (e.g. /tmp/huck/).

Any ideas?

Cheers

eval dir=\$$sourcedir
ls $dir

right, sorry, misunderstood the question...

-Steven