From: Stephane Chazelas <Stephane_Chazelas@yahoo.fr>
To: zsh-users@sunsite.dk
Subject: Re: OT: How to list all but the last item
Date: Mon, 18 Sep 2006 18:17:52 +0100 [thread overview]
Message-ID: <20060918171752.GA4980@sc> (raw)
In-Reply-To: <20060918.183959.74746293.Meino.Cramer@gmx.de>
On Mon, Sep 18, 2006 at 06:39:59PM +0200, Meino Christian Cramer wrote:
> Hi,
>
> may be this is a very stupid question...and may be I am blind...
> But...
>
> I want to contruct a loop like
>
> for i in `<cmd>`
> do
> <do something> ${i}
> done
>
> and <cmd> should return a list of items matched by a regexp
> or another kind of qualifier and skipping the last item.
>
> Example:
>
> ls -rtlc * | <???what???>
>
> would return every item in a directory exept the newest one.
-c is to sort of the file change-status time. You want file
modification time, it's ls -rtl.
And it should be
ls -rtl | ...
Or
ls -rtld -- * | ...
> Is there any way to accomplish with something fitting in on
> a commandline???
ls -tl | tail +2
ls -trl | sed '$d'
Also:
ls -trld -- *(om[2,-1])
Also:
IFS=$'\n\n'
lines=( $(cmd) )
for line in "${(@)lines[1,-2]}"; do ...; done
--
Stéphane
next prev parent reply other threads:[~2006-09-18 17:18 UTC|newest]
Thread overview: 5+ messages / expand[flat|nested] mbox.gz Atom feed top
2006-09-18 16:39 Meino Christian Cramer
2006-09-18 17:02 ` Will Maier
2006-09-18 17:17 ` Meino Christian Cramer
2006-09-18 17:17 ` Stephane Chazelas [this message]
2006-09-18 17:23 ` Meino Christian Cramer
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