bash - Command to get nth line of STDOUT

Question

Is there any bash command that will let you get the nth line of STDOUT?

That is to say, something that would take this

$ ls -l
-rw-r--r--@ 1 root  wheel my.txt
-rw-r--r--@ 1 root  wheel files.txt
-rw-r--r--@ 1 root  wheel here.txt

and do something like

$ ls -l | magic-command 2
-rw-r--r--@ 1 root  wheel files.txt

I realize this would be bad practice when writing scripts meant to be reused, BUT when working with the shell day to day it'd be useful to me to be able to filter my STDOUT in such a way.

I also realize this would be semi-trivial command to write (buffer STDOUT, return a specific line), but I want to know if there's some standard shell command to do this that would be available without me dropping a script into place.

Answer 1

Using sed, just for variety:

ls -l | sed -n 2p

Using this alternative, which looks more efficient since it stops reading the input when the required line is printed, may generate a SIGPIPE in the feeding process, which may in turn generate an unwanted error message:

ls -l | sed -n -e '2{p;q}'

I've seen that often enough that I usually use the first (which is easier to type, anyway), though ls is not a command that complains when it gets SIGPIPE.

For a range of lines:

ls -l | sed -n 2,4p

For several ranges of lines:

ls -l | sed -n -e 2,4p -e 20,30p
ls -l | sed -n -e '2,4p;20,30p'