I have a file called alert.log containing the following:

WARNING: received KRVX_OPER_CANNOT_SUPPORT
knlldmm: gdbnm=CROOP
knlldmm: objn=23793
knlldmm: objv=1
knlldmm: scn=5189816456
knllgobjinfo: MISSING Streams multi-version data dictionary!!!
knlldmm: gdbnm=FDROP
knlldmm: objn=49385
knlldmm: objv=2
knlldmm: scn=5201332536

Is there any way I can use egrep to pull out only lines 6-10 out of the file? I tried using:

egrep -n -i "knllgobjinfo|knlldmm" alert.log

but that also gives me lines 2-5 which I dont want.
If egrep can't do it, what else can I use that can?

Thanks in advance.

You can also use regex ranges:

hi jim.First command is printing 6-10 but the second one is printing only 6-7 lines.

typo, my bad.:
(this assumes there is a \n on the end of the last line)

Thanks a lot, you answered my question, but its not exactly what I was looking, I wasn't as specific as I should have been.

What I really meant is: How can I print to standard output the line beginning with "knllgobjinfo" and its following 4 lines, regardless of where it occurs in the file.

Thanks again.

Dude try following.

sed -n -e '/knllgobjinfo/,+4 p' datafile

Cheers,
Nilesh

nope, this is the error that I get.

sed: 0602-404 Function /knllgobjinfo/,+4 p cannot be parsed.