Re: Sorting directories by size

[Alex fxmbsw7 Ratchev]

i guess you didnt set $dir

On Tue, Aug 24, 2021, 14:12 hancooper <hancooper@protonmail.com> wrote:

> ‐‐‐‐‐‐‐ Original Message ‐‐‐‐‐‐‐
> On Tuesday, August 24, 2021 3:31 AM, Kerin Millar <kfm@plushkava.net>
> wrote:
>
> > On Tue, 24 Aug 2021 02:21:28 +0000
> > hancooper via help-bash@gnu.org wrote:
> >
> > > I have a directory stored in `dir`. I want to store the names of
> directories at a particular level
> > > and sort them by size.
> > > I have started with listing the directory names in `dlc` and computing
> the total number of
> > > directories present. But still have to sort `dlc` by size, the with
> the biggest first.
> > > My initial plan is to use a command such as `du -h | sort -h`.
> > > daggr=( -mindepth "$depth" -maxdepth "$depth" )
> > > dlc=$( find "$dir" "${daggr[@]}" -type d | tr "\n" " " )
> > > ndr=$( echo "$dlc;" | grep -o " " | wc -l )
> >
> > The last two of these three commands should be jettisoned. Further, you
> ought to be using an array to store the directory names. As you appear to
> be using GNU, consider the following.
> >
> > daggr=(-mindepth "$depth" -maxdepth "$depth")
> > mapfile -td '' dlc < <(find "$dir" "${daggr[@]}" -type d -exec du -0 {}
> + | sort -zrn | cut -z -f2-)
> > ndr=${#dlc[@]}
> >
> > This relies on the fact that GNU du(1), given the -0 option, employs an
> output format of "%d\t%s\0", <size>, <pathname>.
> >
> > --
> >
> > Kerin Millar
>
>
> I agree with your assessment.  Have seen that I get the following errer
>
> find: ‘’: No such file or directory
>
>
>