How to print third column to last column?

Question

I\'m trying to remove the first two columns (of which I\'m not interested in) from a DbgView log file. I can\'t seem to find an example that prints from column 3 onwards unt

Answer 1

awk '{ print substr($0, index($0,$3)) }'

solution found here:
http://www.linuxquestions.org/questions/linux-newbie-8/awk-print-field-to-end-and-character-count-179078/

Answer 2

awk '{a=match($0, $3); print substr($0,a)}'

First you find the position of the start of the third column. With substr you will print the whole line ($0) starting at the position(in this case a) to the end of the line.

Answer 3

In Bash you can use the following syntax with positional parameters:

while read -a cols; do echo ${cols[@]:2}; done < file.txt

Learn more: Handling positional parameters at Bash Hackers Wiki

Answer 4

What about following line:

awk '{$1=$2=$3=""; print}' file

Based on @ghostdog74 suggestion. Mine should behave better when you filter lines, i.e.:

awk '/^exim4-config/ {$1=""; print }' file

Answer 5

awk '{$1=$2=""}1' FILENAME | sed 's/\s\+//g'

First two columns are cleared, sed removes leading spaces.

Answer 6

The following awk command prints the last N fields of each line and at the end of the line prints a new line character:

awk '{for( i=6; i<=NF; i++ ){printf( "%s ", $i )}; printf( "\n"); }'

Find below an example that lists the content of the /usr/bin directory and then holds the last 3 lines and then prints the last 4 columns of each line using awk:

$ ls -ltr /usr/bin/ | tail -3
-rwxr-xr-x 1 root root       14736 Jan 14  2014 bcomps
-rwxr-xr-x 1 root root       10480 Jan 14  2014 acyclic
-rwxr-xr-x 1 root root    35868448 May 22  2014 skype

$ ls -ltr /usr/bin/ | tail -3 | awk '{for( i=6; i<=NF; i++ ){printf( "%s ", $i )}; printf( "\n"); }'
Jan 14 2014 bcomps 
Jan 14 2014 acyclic 
May 22 2014 skype