Diff for "BashFAQ/011"

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

One dirty (but not quick) way is:

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

But this reads the entire file even if only the third line is desired, which can be avoided by using the q command to quit on line $n, and deleting all other lines with the d command:

   1 sed "${n}q;d" "$file"

There is yet another simple way of printing only the line you want using sed. The -n option will not print anything unless p flag is explicitly specified. For example to print the 4th line in a file use.

   1 sed -n "4p" "$file"

Another method is to grab lines starting at n, then get the first line of that.

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

Another approach, using AWK:

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

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

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

Or a counter with a simple read loop:

   1 # Bash/ksh
   2 m=0
   3 while ((m++ < n)) && read -r _; do
   4     :
   5 done
   6 
   7 head -n 1

To read into a variable, it is preferable to use read or mapfile rather than an external utility. More than one line can be read into the given array variable or the default array MAPFILE by adjusting the argument to mapfile's -n option:

   1 # Bash4
   2 mapfile -ts "$((n - 1))" -n 1 x <"$file"
   3 printf '%s\n' "$x"

See Also

• BashFAQ/001

• http://wiki.bash-hackers.org/commands/builtin/mapfile

CategoryShell