Bash tool to get nth line from a file

Question

Is there a \"canonical\" way of doing that? I\'ve been using head -n | tail -1 which does the trick, but I\'ve been wondering if there\'s a Bash tool that speci

Answer 1

This question being tagged Bash, here's the Bash (≥4) way of doing: use mapfile with the -s (skip) and -n (count) option.

If you need to get the 42nd line of a file file:

mapfile -s 41 -n 1 ary < file

At this point, you'll have an array ary the fields of which containing the lines of file (including the trailing newline), where we have skipped the first 41 lines (-s 41), and stopped after reading one line (-n 1). So that's really the 42nd line. To print it out:

printf '%s' "${ary[0]}"

---

If you need a range of lines, say the range 42–666 (inclusive), and say you don't want to do the math yourself, and print them on stdout:

mapfile -s $((42-1)) -n $((666-42+1)) ary < file
printf '%s' "${ary[@]}"

If you need to process these lines too, it's not really convenient to store the trailing newline. In this case use the -t option (trim):

mapfile -t -s $((42-1)) -n $((666-42+1)) ary < file
# do stuff
printf '%s\n' "${ary[@]}"

You can have a function do that for you:

print_file_range() {
    # $1-$2 is the range of file $3 to be printed to stdout
    local ary
    mapfile -s $(($1-1)) -n $(($2-$1+1)) ary < "$3"
    printf '%s' "${ary[@]}"
}

No external commands, only Bash builtins!