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Thanks a lot, it's woks perfectly, can you please explain about the following code.
OLDIFS="$IFS"
IFS="|"
set -- $line
IFS="$OLDIFS"

Hi,
i have in a situation to get the variables which are having a "|" delimiter and each variable may or may not have spaces, so how can i skip the spaces on the string and print the string as it...

Excellent!!!
It's working , thanks alot UNIX Forums.

Here is the modified code.
echo "HOSTNAME1:" "$HOSTNAME1" | awk -v var="Everyone" '{printf("%- 12s %- 40.40s %s\n",$1,$2,var)}'
echo...

Thanks for your reply....
I don't know the length of the second field, all data is coming dynamically, is there a way i can print 3 field starting from 50th location towards right, doesn't matter...

Hi,
I am in a situation to print the message on a column, where the each line starting position should be same.

For example code:
HOSTNAME1="1.2.3.4.5.6.7"
TARGET_DIR="/tmp"
echo...