Re: script line not working as its supposed to, but why?

["Gian Uberto Lauri" <saint@eng.it>]

Sharon Kimble writes:
 > But there is no consistency with the creation date, the menu itself is
 > regenerated whenever I install a new programme, and the old menu is
 > saved with the suffix of the date and time.

 > I want to delete the 'menu-*' files if there are more than 7, and
 > the command is parsed when I have 'set -x' at the head of the
 > script but this line does nothing! It runs but doesn't achieve
 > anything.

The find command is not your friend here, I think.

If you have a single directory where these menu are regenerated
automatically, you could exploit the fact that lines like

menu-20131209-11:05

have an asciibetic order (order based on the ASCII code) that matches
the age order, but reversed, I mean newer backups come last. The
option -t of ls fixes this.

I would try with something like this:

if [  `ls -1 menu-* | wc -l` -gt 7 ]
then
    ls -t menu-* | tail $((7-`ls | wc -l`)) | xargs rm
fi

The first test ensures that you have more than 7 files.

Then you list the files in reverse asciibetical order (that is older
last), then the expression

$((7-`ls | wc -l`))

does the magic to compute the option to pass to tail so that it shows
the last (number of files - 7). And finally xargs feeds rm. You can
use rm -v to see them being deleted :)

I hope this helps.

-- 
 /\           ___                                    Ubuntu: ancient
/___/\_|_|\_|__|___Gian Uberto Lauri_____               African word
  //--\| | \|  |   Integralista GNUslamico            meaning "I can
\/                 coltivatore diretto di software       not install
     già sistemista a tempo (altrui) perso...                Debian"

Warning: gnome-config-daemon considered more dangerous than GOTO