BashFAQ/011

How can I print the n'th line of a file?

One dirty (but not quick) way is:

    sed -n ${n}p "$file"

But this reads the entire file even if only the third line is desired, which can be avoided by printing line $n using the "p" command, followed by a "q" to exit the script:

    sed -n "$n{p;q;}" "$file"

Another method is to grab the last line from a listing of the first n lines:

   head -n $n $file | tail -n 1

Another approach, using AWK:

   awk "NR==$n{print;exit}" file

If more than one line is needed, it's easy to adapt any of the previous methods:

   x=3 y=4
   sed -n "$x,${y}p;${y}q;" "$file"                # Print lines $x to $y; quit after $y.
   head -n $y "$file" | tail -n $((y - x + 1))     # Same
   head -n $y "$file" | tail -n +$x                # If your tail supports it
   awk "NR>=$x{print} NR==$y{exit}" "$file"        # Same

In Bash 4, a pure-bash solution can be achieved succinctly using the mapfile builtin. More than one line can be read into the array MAPFILE by adjusting the argument to mapfile's -n option:

   mapfile -ts $((n-1)) -n 1 < "$file"
   echo "${MAPFILE[0]}"

mapfile can also be used similarly to head while avoiding buffering issues in the event input is a pipe:

   { mapfile -n $n; head -n 1; } <"$file"

1. See Also

CategoryShell