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

User · Answer

Try this sed version   ls -l   sed  2   d    It says  delete all the lines that aren t the second one

User · Answer

ls -l   head -2   tail -1

User · Answer

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

User · Answer

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

User · Answer

Is Perl easily available to you     perl -n -e  if        7    print  exit 0        Obviously substitute whatever number you want for 7

User · Answer

Alternative to the nice head   tail way   ls -al   awk  NR  2    or  ls -al   sed -n  2p

User · Answer

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

User · Answer

Yes  the most efficient way  as already pointed out by Jonathan Leffler  is to use sed with print  amp  quit   set -o pipefail                          cf  help set time -p ls -l   sed -n -e  2 p q         only print the second line  amp  quit  on Mac OS X  echo        PIPESTATUS                   cf  man bash   less -p  PIPESTATUS

User · Answer

From sed1line     print line number 52 sed -n  52p                    method 1 sed  52 d                      method 2 sed  52q d                     method 3  efficient on large files   From awk1line     print line number 52 awk  NR  52  awk  NR  52  print exit              more efficient on large files

User · Answer

For the sake of completeness  -   shorter code  find     awk NR  3   shorter life  find     awk  NR  3  print  0  exit

User · Answer

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

User · Answer

For more completeness    ls -l    for   x 0 x lt 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

[bash] Command to get nth line of STDOUT

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