Very Computer

hello,

I have a file that appears to be delimited by three spaces between
fields.   I tried using cut to seperate the fields, but cut will only
take single character delimiters.   What I need to do is extract
columns 2,3, and 8, then sort the columns based on column 2.   There's
also a file header, but I don't want to keep it...   Is there a way to
do this from the bash prompt?

thanks.

        awk '{print $2,$3,$8}' file

or this:

        while read line
        do
                set -- "$line"
                echo "$2 $3 $8"
        done < file

will give you the columns you want. If you need to work with exactly 3
spaces as the field separator, this will do it:

        gawk -F"   " '{print $2,$3,$8}' file

In either case, then just use "sort" as normal.

        Ed.

With tr: ( that is a quote spacebar quote )
ls -l | tr -s ' '
drwxr-xr-x 2 jbenn02 users 48 2005-06-21 08:27 bin
-rw-r--r-- 1 jbenn02 users 1047 2005-07-07 19:26 callcy.txt

Now I can cut with ease:
ls -l |tr -s ' ' |cut -d ' ' -f8
bin
callcy.txt

It sounds like then you should be able to pipe your info to sort.
Example lines ??

Ponders the JB.

{
  read ## Discard header (assuming it's just the first line)
  awk '{print $2, $3, $8}'

Quote:} < FILENAME | sort

{
  read ## Discard header (assuming it's just the first line)
  sed "s/   /\t/g" | awk 'BEGIN {FS = "\t"} {print $2, $3, $8}'

Quote:} < FILENAME | sort