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

Question

right now I have this line, and it worked until I had whitespace in the second field.

svn status | grep '\!' | gawk '{print $2;}' > removedProjs

is there a way to have awk print everything in $2 or greater? ($3, $4.. until we don't have anymore columns?)

I suppose I should add that I'm doing this in a windows environment with cygwin.

Answer 1

You could use a for-loop to loop through printing fields $2 through $NF (built-in variable that represents the number of fields on the line).

Edit: Since "print" appends a newline, you'll want to buffer the results:

awk '{out=""; for(i=2;i<=NF;i++){out=$out" "$i}; print $out}'

Alternatively, use printf:

awk 'for(i=2;i<=NF;i++){printf "%s ", $i}; printf "\n"}'

Answer 2

Would this work?

awk '{print substr($0,length($1)+1);}' < file

It leaves some trailing whitespace in front though.

Answer 3

will print all but very first column:

cat somefile | awk '{$1=""; print $0}'    

will print all but two first columns:

cat somefile | awk '{$1=$2=""; print $0}'