using awk script to print last n rows of a text file

Question

I need a script to print last n rows of a text file. the text file names and number of rows can be varied and I want to call only script to print last n rows of any text files. I know for first n row I can use NR < n; print but how can I do it for last n row as number of rows can be varied.- thanks

Answer 1

There is a unix tool for exactly that purpose, called tail. To get the last 100 rows of file, you would use tail -n 100 file, then use the output directly or pipe it to other programs like awk.

Answer 2

To do this natively in awk, you have to remember the lines as you see them:

awk -v n=10 '
    {line[NR]=$0}
    END {for (i=NR-(n-1); i<=NR; i++) print line[i]}
' file

To save memory, we don't need to remember the whole file; use

    {line[NR]=$0; if (NR>n) delete line[NR-n]}

However it is simpler to reverse the file, print the first n lines, and re-reverse the output

tac file | awk -v n=10 'NR <= n' | tac

But using tail is much simpler that all that