awk / cut: Skip First Two Fields and Print the Rest of Line

by nixCraft on July 13, 2011 · 7 comments· LAST UPDATED July 13, 2011

in BASH Shell

I'd like to skip first two or three fields at the the beginning of a line and print the rest of line. Consider the following input:

This is a test
Giving back more than we take

I want my input file with the following output:

a test
more than we take

How do I printing lines from the nth field using awk under UNIX or Linux operating systems?

You can use the awk command as follows:

 
echo 'This is a test' | awk '{print substr($0, index($0,$3))}'
 

OR

 
awk '{print substr($0, index($0,$3))}' <<< 'This is a test'
 

You can also use the cut command:

 
echo 'This is a test' | cut -d ' ' -f3-
 

OR

 
cut -d ' ' -f3- <<<'This is a test'
 

Finally, process the file using bash while loop:

#!/bin/bash
_input="/path/to/file.name.txt"
while IFS= read -r line
do
    cut -d ' ' -f3- <<<"$line"
    ### same stuff with awk ###
    ### awk '{print substr($0, index($0,$3))}' <<< "$line" ###
done < "${_input}"