Print second last column/field in awk

Question

I want to print the second last column or field in awk. The number of fields is variable. I know that I should be able to use $NF but not sure how it can be use

Answer 1

Did you tried to start from right to left by using the rev command ? In this case you just need to print the 2nd column:

seq 12 | xargs -n5 | rev | awk '{ print $2}' | rev
4
9
11

Answer 2

awk '{print $(NF-1)}'

Should work

Answer 3

awk ' { print ( $(NF-1) ) }' file