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

There's a duplicate question with a simpler answer using cut:

 svn status |  grep '\!' | cut -d\  -f2-

-d specifies the delimeter (space), -f specifies the list of columns (all starting with the 2nd)

Print all columns:

awk '{print $0}' somefile

Print all but the first column:

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

Print all but the first two columns:

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

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"}'

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