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 until the end of the line. Note that each line has variable number of columns.

Answer 1

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

Answer 2

A bit late here, but none of the above seemed to work. Try this, using printf, inserts spaces between each. I chose to not have newline at the end.

awk '{for(i=3;i<=NF;++i) printf("%s ",  $i) }'

Answer 3

awk -v m="\x0a" -v N="3" '{$N=m$N ;print substr($0, index($0,m)+1)}'

This chops what is before the given field nr., N, and prints all the rest of the line, including field nr.N and maintaining the original spacing (it does not reformat). It doesn't mater if the string of the field appears also somewhere else in the line, which is the problem with daisaa's answer.

Define a function:

fromField () { 
awk -v m="\x0a" -v N="$1" '{$N=m$N; print substr($0,index($0,m)+1)}'
}

And use it like this:

$ echo "  bat   bi       iru   lau bost   " | fromField 3
iru   lau bost   
$ echo "  bat   bi       iru   lau bost   " | fromField 2
bi       iru   lau bost 

Output maintains everything, including trailing spaces

Works well for files where '/n' is the record separator so you don't have that new-line char inside the lines. If you want to use it with other record separators then use:

awk -v m="\x01" -v N="3" '{$N=m$N ;print substr($0, index($0,m)+1)}'

for example. Works well with almost all files as long as they don't use hexadecimal char nr. 1 inside the lines.

Answer 4

awk '{for (i=4; i<=NF; i++)printf("%c", $i); printf("\n");}'

prints records starting from the 4th field to the last field in the same order they were in the original file

Answer 5

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 6

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.