Diff for "BashFAQ/011"

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

The dirty (but not quick) way would be:

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

but this reads the whole input file, even if you only wanted the third line.

This one avoids that problem:

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

At line $n the command "p" is run, printing it, with a "q" afterwards: quit the program.

Another way, more obvious to some, 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 you want more than one line, it's pretty 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

Note

In most cases, you should sanitize your variable n to be sure, that it's not containing any of non-digits, before feeding it to sed or awk. You can do it with such simple code:

   # Bash
   n=${n//[!0-9]/}

   # POSIX
   n=$(printf "%s" "$n"|tr -cd '0-9')

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