Is there a way to ignore header lines in a UNIX sort?

Question

I have a fixed-width-field file which I\'m trying to sort using the UNIX (Cygwin, in my case) sort utility.

The problem is there is a two-line header at the top of t

Answer 1

cat file_name.txt | sed 1d | sort 

This will do what you want.

Answer 2

Here is a version that works on piped data:

(read -r; printf "%s\n" "$REPLY"; sort)

If your header has multiple lines:

(for i in $(seq $HEADER_ROWS); do read -r; printf "%s\n" "$REPLY"; done; sort)

This solution is from here

Answer 3

It only takes 2 lines of code...

head -1 test.txt > a.tmp; 
tail -n+2 test.txt | sort -n >> a.tmp;

For a numeric data, -n is required. For alpha sort, the -n is not required.

Example file:
$ cat test.txt

header
8
5
100
1
-1

Result:
$ cat a.tmp

header
-1
1
5
8
100

Answer 4

This is the same as Ian Sherbin answer but my implementation is :-

cut -d'|' -f3,4,7 $arg1 | uniq > filetmp.tc
head -1 filetmp.tc > file.tc;
tail -n+2 filetmp.tc | sort -t"|" -k2,2 >> file.tc;

Answer 5

You can use tail -n +3 <file> | sort ... (tail will output the file contents from the 3rd line).

Answer 6

If you don't mind using awk, you can take advantage of awk's built-in pipe abilities

eg.

extract_data | awk 'NR<3{print $0;next}{print $0| "sort -r"}' 

This prints the first two lines verbatim and pipes the rest through sort.

Note that this has the very specific advantage of being able to selectively sort parts of a piped input. all the other methods suggested will only sort plain files which can be read multiple times. This works on anything.