#!/bin/ksh
while read line < elig_jobs.txt
do
#Gets the field from the elig_jobs.txt file that has the input location path.
INPUTD=`echo "$line" | cut -c240-289` (ex: $HOME/2005)

echo inputdirectory: $INPUTD (this prints $HOME/2005)

I want it to print /data/user/2005 rather than $HOME/2005/ Could some body please see what's the problem here. I even tried with double quotes:
echo inputdirectory: "$INPUTD" it still prints $HOME/2005


ThankYou,
Radhika.

it would be nice to see a couple of lines from 'elig_jobs.txt' file......

elig_jobs.txt file is a file with some data about a .dat file-

Row 1-------------------------
Eligibility 1 120050320050516122117xxx121_xxxxx_tgt_bn*.dat
gsk_target_bonus.txt $HOME/2005/ \\serverxxxx\ssd_sim\Data

Row 2-------------------------
Eligibility 2 220050320050516122117xxx121_xxxxx_sta*.dat
gsk_rep_status.txt $HOME/2005/ \\serverxxxx\ssd_sim\Data


For your testing purposes I think it would be simple to consider a simple text file with
$HOME/2005 as data in one row.
And use
INPUTD=`echo "$line" | cut -c1-10` (ex: $HOME/2005) to clip the path from the file.

#!/bin/ksh

while read line
do
   INPUTD=`eval echo "$line" | cut -d ' ' -f5`

   echo inputdirectory: $INPUTD
done < elig_jobs.txt

Thank You!!! so much. It works nicely.

Radhika.