Using awk to print all columns from the nth to the last

Most solutions with awk leave an space. The options here avoid that problem.

Option 1

A simple cut solution (works only with single delimiters):

command | cut -d' ' -f3-

Option 2

Forcing an awk re-calc sometimes remove the added leading space (OFS) left by removing the first fields (works with some versions of awk):

command | awk '{ $1=$2="";$0=$0;} NF=NF'

Option 3

Printing each field formatted with printf will give more control:

$ in=' 1 2 3 4 5 6 7 8 '
$ echo "$in"|awk -v n=2 '{ for(i=n+1;i<=NF;i++) printf("%s%s",$i,i==NF?RS:OFS);}'
3 4 5 6 7 8

However, all previous answers change all repeated FS between fields to OFS. Let's build a couple of option that do not do that.

Option 4 (recommended)

A loop with sub to remove fields and delimiters at the front. And using the value of FS instead of space (which could be changed).

Is more portable, and doesn't trigger a change of FS to OFS: NOTE: The ^[FS]* is to accept an input with leading spaces.

$ in=' 1 2 3 4 5 6 7 8 '
$ echo "$in" | awk '{ n=2; a="^["FS"]*[^"FS"]+["FS"]+"; for(i=1;i<=n;i++) sub( a , "" , $0 ) } 1 '
3 4 5 6 7 8

Option 5

It is quite possible to build a solution that does not add extra (leading or trailing) whitespace, and preserve existing whitespace(s) using the function gensub from GNU awk, as this:

$ echo ' 1 2 3 4 5 6 7 8 ' | awk -v n=2 'BEGIN{ a="^["FS"]*"; b="([^"FS"]+["FS"]+)"; c="{"n"}"; } { print(gensub(a""b""c,"",1)); }'
3 4 5 6 7 8 

It also may be used to swap a group of fields given a count n:

$ echo ' 1 2 3 4 5 6 7 8 ' | awk -v n=2 'BEGIN{ a="^["FS"]*"; b="([^"FS"]+["FS"]+)"; c="{"n"}"; } { d=gensub(a""b""c,"",1); e=gensub("^(.*)"d,"\\1",1,$0); print("|"d"|","!"e"!"); }'
|3 4 5 6 7 8 | ! 1 2 !

Of course, in such case, the OFS is used to separate both parts of the line, and the trailing white space of the fields is still printed.

NOTE: [FS]* is used to allow leading spaces in the input line.