How to print all the columns after a particular number using awk?

Question

On shell, I pipe to awk when I need a particular column.

This prints column 9, for example:

... | awk '{print $9}'

How can I tell awk to print all the columns including and after column 9, not just column 9?

Answer 1

awk '{ s = ""; for (i = 9; i <= NF; i++) s = s $i " "; print s }'

Answer 2

When you want to do a range of fields, awk doesn't really have a straight forward way to do this. I would recommend cut instead:

cut -d' ' -f 9- ./infile

Edit

Added space field delimiter due to default being a tab. Thanks to Glenn for pointing this out

Answer 3

awk '{print substr($0, index($0,$9))}'

Edit: Note, this doesn't work if any field before the ninth contains the same value as the ninth.

Answer 4

sed -re 's,\s+, ,g' | cut -d ' ' -f 9-

Instead of dealing with variable width whitespace, replace all whitespace as single space. Then use simple cut with the fields of interest.

It doesn't use awk so isn't germane but seemed appropriate given the other answers/comments.

Answer 5

Generally perl replaces awk/sed/grep et. al., and is much more portable (as well as just being a better penknife).

perl -lane 'print "@F[8..$#F]"'

Timtowtdi applies of course.

Answer 6

awk -v m="\x01" -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 Ascherer's answer.

Define a function:

fromField () { 
awk -v m="\x01" -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 For N=0 it returns the whole line, as is, and for n>NF the empty string

Answer 7

Here is an example of ls -l output:

-rwxr-----@ 1 ricky.john  1493847943   5610048 Apr 16 14:09 00-Welcome.mp4
-rwxr-----@ 1 ricky.john  1493847943  27862521 Apr 16 14:09 01-Hello World.mp4
-rwxr-----@ 1 ricky.john  1493847943  21262056 Apr 16 14:09 02-Typical Go Directory Structure.mp4
-rwxr-----@ 1 ricky.john  1493847943  10627144 Apr 16 14:09 03-Where to Get Help.mp4

My solution to print anything post $9 is awk '{print substr($0, 61, 50)}'

Answer 8

To display the first 3 fields and print the remaining fields you can use:

awk '{s = ""; for (i=4; i<= NF; i++) s= s $i : "; print $1 $2 $3 s}' filename

where $1 $2 $3 are the first 3 fields.

Answer 9

function print_fields(field_num1, field_num2){
    input_line = $0

    j = 1;
    for (i=field_num1; i <= field_num2; i++){
        $(j++) = $(i);

    }
    NF = field_num2 - field_num1 + 1;
    print $0

    $0 = input_line
}

Answer 10

Using cut instead of awk and overcoming issues with figuring out which column to start at by using the -c character cut command.

Here I am saying, give me all but the first 49 characters of the output.

 ls -l /some/path/*/* | cut -c 50-

The /*/*/ at the end of the ls command is saying show me what is in subdirectories too.

You can also pull out certain ranges of characters ala (from the cut man page). E.g., show the names and login times of the currently logged in users:

       who | cut -c 1-16,26-38

Answer 11

ruby -lane 'print $F[3..-1].join(" ")' file