Most solutions with awk leave a 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 options 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).
This 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 preserves 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 whitespace of the fields is still printed.
NOTE: [FS]*
is used to allow leading spaces in the input line.
grep | awk
is an antipattern -- you wantawk '/!/ { print $2 }'
svn status | grep '\!' | cut -d' ' -f2- > removedProjs