Taking x number of last columns from a file in awk - bash

Question

I have a txt file with columns such as name, position, department, etc. I need to get certain characters from each column to generate a username and password for each listing.
I'm approaching this problem by first separating the columns with:
awk 'NR>10{print $1, $2, $3, ...etc}' $filename
#This skips the header information and simply prints out each specified column.
And then moving on to grab each character that I need from their respective fields.
The problem now is that each row is not uniform. For example the listings are organized as such:
(lastName, firstName, middleInitial, position, departmentName)
Some of the entries in the text file do not have certain fields such as middle initial. So when I try to list fields 3, 4, and 5 the entries without their middle initials return:
(position, departmentName, empty/null (I'm not sure how awk handles this))
The good news is that I do not need the middle initial, so I can ignore these listings . How can I go about grabbing the last 2 (in this example) columns from the file so I can isolate the fields that every entry has in order to cut the necessary characters out of them?

Answer 1

You can get them by $(NF-1) and $NF, they are last 2nd column and last column
echo "1,2,3,4" | awk -F, 'BEGIN{OFS=","}{print $(NF-1), $NF}'
NF means number of fields. If you have 4 fields. (NF-1) would be column 3, and $(NF-1) is the value of column 3.
Output would be
3,4
Another example with different length fields in file:
sample.csv
1,2,3,4,5
a,b,c,d
Run
awk -F, 'BEGIN{OFS=","}{print $(NF-1), $NF}' sample.csv
Output:
4,5
c,d