Shell Programming and Scripting

Hi All,

I need to print the lines after 2nd line irrespective of the number of lines in file

if the file contents are like below
----------
root:!:0:0::/root:/usr/bin/ksh
daemon:!:1:1::/etc:
bin:!:2:2::/bin:
sys:!:3:3::/usr/sys:
adm:!:4:4::/var/adm:
uucp:!:5:5::/usr/lib/uucp:
guest:!:100:100::/home/guest:
nobody:!:4294967294:4294967294::/:
lpd:!:9:4294967294::/:
.
.
.

I need from the line after "daemon" line like as below
-----
bin:!:2:2::/bin:
sys:!:3:3::/usr/sys:
adm:!:4:4::/var/adm:
uucp:!:5:5::/usr/lib/uucp:
guest:!:100:100::/home/guest:
nobody:!:4294967294:4294967294::/:
lpd:!:9:4294967294::/:
.
.
.

One way,
read the more man page. there's an option to start reading from certain line number.

there are many ways to do this...
I normally use sed for this,

sed -n '3,/ArbitWord/p' <inputFile>

here ArbitWord is something which you can not expect to be present in your input file

one other way is using tac

tac <inputFile> | head --lines=-2 | tac

Or tail +3 /etc/passwd.

The tail command is better in this scenario IMHO, but for completeness, you can use $ (meaning last line of file) instead of /ArbitWord/ in the sed script.

If the line number is unknown and you want to use a search pattern:

Code:

awk '/^daemon/{p=1}p' file

Thank you for all for ur replays
tail +3 worling for my purpose