Date: Sat, 3 Aug 2002 02:58:50 +0200 From: Roman Neuhauser <neuhauser@bellavista.cz> To: freebsd-questions@FreeBSD.ORG Subject: Re: smart file gobbling: exclude one file from list Message-ID: <20020803005850.GR52563@freepuppy.bellavista.cz> In-Reply-To: <20020803002236.GA4398@dan.emsphone.com> References: <001201c23a6e$17e78e60$0200a8c0@capm> <20020802215617.GA9408@dan.emsphone.com> <20020802220620.GN52563@freepuppy.bellavista.cz> <20020803002236.GA4398@dan.emsphone.com>
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> Date: Fri, 2 Aug 2002 19:22:36 -0500
> From: Dan Nelson <dnelson@allantgroup.com>
> To: Pascal Giannakakis <capm@gmx.net>, freebsd-questions@FreeBSD.ORG
> Subject: Re: smart file gobbling: exclude one file from list
>
> In the last episode (Aug 03), Roman Neuhauser said:
> > > In the last episode (Aug 02), Pascal Giannakakis said:
> > > > if i do "cp * /destination", all files in the current directory
> > > > will be copied to /destination. What do i have to type, if i want
> > > > to expand the '*' to all files, except of one? For example, if i
> > > > wanted to exclude the file "notme" in the current directory?
> > > >
> > > > I know i could hack it with grep, but this fails as all files are
> > > > on one line. I wonder how you gurus would solve this! :)
> > >
> > > If you are using zsh and have the EXTENDED_GLOB option set,
> > >
> > > cp *~notme /destination
> > >
> > > Otherwise, you'll need to do something like
> > >
> > > cp $(echo ' ' * ' ' | sed -e 's/ notme / /') /destination
> >
> > this is somewhat easier:
> >
> > cp $(echo `ls|grep -xv notme`) /destination
>
> Although I don't think either will handle files with spaces in them.
mine doesn't.
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