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  wh         


        

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'

Answer 2

Another poster suggested

ls -l | head -2 | tail -1

but if you pipe head into tail, it looks like everything up to line N is processed twice.

Piping tail into head

ls -l | tail -n +2 | head -n1

would be more efficient?

Answer 3

ls -l | head -2 | tail -1

Answer 4

Is Perl easily available to you?

$ perl -n -e 'if ($. == 7) { print; exit(0); }'

Obviously substitute whatever number you want for 7.

Answer 5

Alternative to the nice head / tail way:

ls -al | awk 'NR==2'

or

ls -al | sed -n '2p'

Answer 6

Hmm

sed did not work in my case. I propose:

for "odd" lines 1,3,5,7... ls |awk '0 == (NR+1) % 2'

for "even" lines 2,4,6,8 ls |awk '0 == (NR) % 2'