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Editor: geirha
Comment: That is just a minor variation of the very first example
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Another method is to grab the last line from a listing of the first `n` lines. However, if the file has less than `n` lines, this will output the last line, unlike the other approaches which will have empty output. Another method is to grab lines starting at `n`, then get the first line of that.
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head -n "$n" "$file" | tail -n 1 tail -n "+$n" "$file" | head -n 1

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"

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

CategoryShell

BashFAQ/011 (last edited 2020-05-07 08:35:17 by intranet)