Does anybody know if it if possible to sort a file excluding the first row (Header) or do I have to tail the remaining rows to another file and then sort in that file. If this is so, I need a little further help with awk and NR to utilise the number of rows, and then subtract by 1 to tail the file. I've nearly got it but just can't get it to work.

I don't know about skipping a one line header but ((line_count = `wc -l filename | awk '{ print $1 }` - 1)) should set line_count to one less than the length of the file.

Thanks helped a lot....

Using David's example, and taking it a bit further, you should be able to: #!/bin/ksh ((line_count = `wc -l filename | awk '{ print $1 }` - 1)) tail -$line_count filename | sort

"tail -nnn" will go back only so far, so this solution fails on larger files. To skip the first part of a file, you don't need to know the line count: tail +2 filename | sort

Quite nice to know James. Learning something new every day. Thanks! if thats the case then, why need the line count of the file? Why not simply use the tail +2 filename|sort ?.

Yep, that is my solution. Like I said, you do not need the line count to skip the first n lines of a file.