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

shell awk

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

shell awk

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

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

shell awk

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.

CodeHunter