Diff for "BashFAQ/011"

Differences between revisions 37 and 38

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"

-  ⇤ ← Revision 37 as of 2018-12-21 20:48:17 → 
  Size: 1920
  Editor: Valentin Bajrami
  Comment:
+   ← Revision 38 as of 2018-12-23 22:43:52 → ⇥
  Size: 1658
  Editor: geirha
  Comment: That is just a minor variation of the very first example
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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.

{{{#!highlight bash
sed -n "4p" "$file"

Diff for "BashFAQ/011"

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

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