Re: df -h listing
On Mon, Oct 08, 2007 at 08:18:02AM -0400, Rick Pasotto wrote:
> On Mon, Oct 08, 2007 at 08:12:33AM -0400, Kevin Mark wrote:
> > On Mon, Oct 08, 2007 at 01:46:10PM +0200, Wolodja Wentland wrote:
> > > On Mon, 2007-10-08 at 13:24 +0200, roberto wrote:
> > > > hello
> > > > i use the "df -h" or "du -h"command to check how much disk space is
> > > > occupied by files and directories
> > > >
> > > > but is it possible to sort the output list in an order such that the
> > > > first (or conversely the last) item is the largest in size ?
> > >
> > > You can achieve this by sorting the output produced by df/du like this:
> > >
> > > "df -h|sort -k2"
> > >
> > > The "-k2" tells sort to sort by the second column. If you want it sorted
> > > in reverse order add the "-r" option.
> > >
> > > Hope that answers your question
> > I think you missed someting? like the -n flag? If you do a normal sort,
> > its alphabetic. With -n, it is done numeric.
>
> That's true but it doesn't help anyway. 57K will sort larger than 2M.
I forgot to mention that I never use the '-h' option for this reason and
use '-m'. What use is the 'human' size, when you want to sort numbers by
machine, thus you need similar units, unless the sort routine can
understnd the -h notation, and unix tools usually avoid this unneeded
complexity.
-K
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