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.
magic-command
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'
ls -l | head -2 | tail -1
head -n 2 | tail -n 1
— modern heads and tails tend to issue warnings otherwise.
head -n M | tail -n M-N+1
), and generalizing to handle multiple ranges is not possible.
$ ls -l | awk '{if (NR>=4 && NR<=64) print}'
(can add more conditions easier, multiple ranges, etc...)
Alternative to the nice head / tail way:
ls -al | awk 'NR==2'
or
ls -al | sed -n '2p'
'2p'
stand for?
2
is a variable? Every example uses single quotes, but with a var you need double quotes. Could we "program" awk to work also with single quotes when using vars? Example line =1; ls | awk 'NR==$line'
. I know that bash
uses double quotes with vars.
For the sake of completeness ;-)
shorter code
find / | awk NR==3
shorter life
find / | awk 'NR==3 {print $0; exit}'
ls
— the original question was about someone's STDOUT
, so I thought it's better to have it bigger.
{ print $0 }
, so `awk 'NR==3' is a shorter way to write the same.
Try this sed
version:
ls -l | sed '2 ! d'
It says "delete all the lines that aren't the second one".
You can use awk:
ls -l | awk 'NR==2'
Update
The above code will not get what we want because of off-by-one error: the ls -l command's first line is the total line. For that, the following revised code will work:
ls -l | awk 'NR==3'
Is Perl easily available to you?
$ perl -n -e 'if ($. == 7) { print; exit(0); }'
Obviously substitute whatever number you want for 7.
$ awk 'NR % 7' filename
to print every 7th line in file.
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?
Yes, the most efficient way (as already pointed out by Jonathan Leffler) is to use sed with print & quit:
set -o pipefail # cf. help set
time -p ls -l | sed -n -e '2{p;q;}' # only print the second line & quit (on Mac OS X)
echo "$?: ${PIPESTATUS[*]}" # cf. man bash | less -p 'PIPESTATUS'
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'
For more completeness..
ls -l | (for ((x=0;x<2;x++)) ; do read ; done ; head -n1)
Throw away lines until you get to the second, then print out the first line after that. So, it prints the 3rd line.
If it's just the second line..
ls -l | (read; head -n1)
Put as many 'read's as necessary.
Success story sharing
p
is thesed
statement that prints the lines identified by the range of lines that precedes it. Thus2,4p
means print lines 2, 3, 4.n
means to NOT print all the other lines