Shell Programming and Scripting

i having problems figuring out how to 'read' in 3 different directory listings and then display them on the screen into 3 separate columns.

i thought i could use a 'for' loop to grab each directory and then assign a unique variable to each line 'read' in.

of course, as you experts all know, you can't do env$num=$env.

an example of the script i had in mind is below.


#!/bin/ksh
num=0
for env in `ls /mg_apps/crm` `ls /mg_apps/ifid` `ls /mg_apps/riab`
do
env$num=$env
num=`expr $num + 1`
done

echo "crm" "ifid" "riab"
echo $env1 $env4 $env7
echo $env2 $env5 $env8
echo $env3 $env6 $env9



however, this method, even if it did work, also has it's own problems - what happens when a new directory is created? it would throw the variable numbering out and i'd get a 'crm' directory appearing in the 'ifid' column.

all solutions greatly appreciated, i love to see diversity, it increases my knowledge of unix.

cheers.

Code:

ls /mg_apps/crm > /crm_fl
ls /mg_apps/ifid > /fd_fl
ls /mg_apps/riab > /iab_fl
paste /crm_fl /fd_fl /iab_fl

thanks for that. it works, but i didn't want to mess about with creating files and then deleting them.

i also then have to figure out how to get a title at the top of each column and then display in lovely straight lines.

e.g.

crm ifid

crmuat1 ifiduat
crmproduction ifiduat2
crmuat2 ifiddev

actually that hasn't quite displayed correctly. imagine all the columns are lined up.

With bash/ksh93:

Code:

set crm ifid riab
{  printf "%s\t" "$@";printf "\n";eval paste $(printf "<(ls %s) " "$@");}

thank you radoulov, but still not quite what i was after.

once i've placed the full path name of crm, ifid or riab in the 'set' command, i get the full path name as a heading, which is a little messy.

it's also still not aligned in nice straight columns. see the output below.

e.g.

/mg_apps/crm /mg_apps/ifid /mg_apps/riab
crm_MKS_build ifiddev dev
crmdev ifidprd prod
crmprd ifiduat uat
crmuat
reladmin

Code:

{  printf "%s\t" "${@##*/}";printf "\n";eval paste $(printf "<(ls %s) " "$@");}