How to print the nth line of a file?

Easy to write:

sed -n 123456p

Faster, especially if the line is not near the end:

sed '123456q; d'

A bit faster, at least if the file is large ( benchmarks ):

tail -n +123457 | head -n 1

head … | tail …it's much slower because the first 123465 lines go through the pipe.

sedYou can do it using the (Stream Editor) command :

sed 'LINEAq;d' file

Example:

$ cat hello.go
package main

import (
    "fmt"
)

func main() {
    fmt.Printf("hello, world\n")
}

Print the first line:

$ sed '1q;d' hello.go
package main}

Print the seventh line:

$ sed '7q;d' hello.go
func main() {

For more information about sed, you can visit the official GNU documentation.

Several options occur to me.

With awkwe can say:

awk 'NR==n' archivo

and if the number is contained in a variable $numero, we can say:

awk -v linea="$numero" 'NR==linea' archivo

Since NRit refers to the line that is being read at a given moment, we only need to compare its value with that of the variable that we assign to it through -v linea="n". For example, -v linea=5.

If it turns out that the file is very, very large we can always exit after the line has been printed so as not to continue reading it unnecessarily:

awk 'NR==linea {print; exit}' archivo

If we prefer to use sed, we can say:

sed -n 'Np' archivo

And for large files:

sed -n 'N{p;q;}' archivo

This will print line number Nand then stop execution.

If the value is in a variable, we must use double quotes so that the value is expanded and ${n}so that it is understood that the name of the variable is $nand not $np:

sed -n "${n}p" archivo

headand pipewith tailwould be slow for a very large file. I would recommend using sedit like this:

sed 'NUMq;d' archivo

Where NUMis the line number you want to print.

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