Write a program that read from a file and print the line with line number

Question

I need to write a program that will read line from a file and then output the lines.

So the files contains this:

This is the first line
This is the second line
This is the third line
This was a blank line

The output should be this:

1. This is the first line
2. This is the second line
3. This is the third line
4.
5. This was a blank line

I know I can do:

nl -b a tst16

But this doesn't print the "." after the numbers plus I want to know if there is a way to do this is like a loop or something.

• steeldriver: At least in the GNU coreutils version of nl, it seems to be possible to customize the separator string in the way you want e.g. (using bash) nl -b a -s $'.\t' tst16
• MAQ: Try : perl -ne 'print "$.. $_"' file

Answer 1

Using a short while construct:

% i=1; while IFS= read -r line; do printf '%s. %s\n' "$i" "$line"; ((i++)); done <file.txt
1. This is the first line
2. This is the second line
3. This is the third line
4. 
5. This was a blank line

Expanded:

#!/usr/bin/env bash
i=1
while IFS= read -r line; do 
 printf '%s. %s\n' "$i" "$line"
 ((i++))
done <file.txt

Answer 2

You could use nl too --

nl -s". " -ba file
 1. This is the first line
 2. This is the second line
 3. This is the third line
 4. 
 5. This was a blank line

Answer 3

Using awk it is enough to do:

awk ' { print NR". " 1ドル } ' < file.txt

1ドル is the string of text in the input, and NR an internal awk variable that tracks the current line number (e.g. Number of Records).

8 Powerful Awk Built-in Variables – FS, OFS, RS, ORS, NR, NF, FILENAME, FNR