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From: Melvin on 2 Feb 2010 01:07
Hi,

I want my shell to input a file using $argv[1] and cut the .gz part
in it.i.e. if the input file name taken by $argv[1] is xyz.gz, I want
$argv[1] to be assigned as xyz.

How can I do this using cut command.

Thanks in advance
Unix Baby
From: Medernach on 2 Feb 2010 02:58
Melvin wrote:
> Hi,
>
> I want my shell to input a file using $argv[1] and cut the .gz part
> in it.i.e. if the input file name taken by $argv[1] is xyz.gz, I want
> $argv[1] to be assigned as xyz.
>
> How can I do this using cut command.
>
> Thanks in advance
> Unix Baby

You could use the basename command like this:
# basename a.gz .gz
a

Regards,
--
Medernach
From: Chris F.A. Johnson on 2 Feb 2010 03:34
On 2010-02-02, Melvin wrote:
>
> I want my shell to input a file using $argv[1] and cut the .gz part
> in it.i.e. if the input file name taken by $argv[1] is xyz.gz, I want
> $argv[1] to be assigned as xyz.

Where does $argv[1] come from?

> How can I do this using cut command.

If the filename is in a variable, use parameter expansion:

file=xyz.gz
base=${file%.*}


--
Chris F.A. Johnson, author <http://shell.cfajohnson.com/>
===================================================================
Shell Scripting Recipes: A Problem-Solution Approach (2005, Apress)
Pro Bash Programming: Scripting the GNU/Linux Shell (2009, Apress)
===== My code in this post, if any, assumes the POSIX locale =====
===== and is released under the GNU General Public Licence =====
From: Melvin on 2 Feb 2010 03:58
On Feb 2, 12:58 pm, Medernach <medern...(a)clermont.in2p3.fr> wrote:
> Melvin wrote:
> > Hi,
>
> > I want my shell to input a file using $argv[1] and cut the .gz part
> > in it.i.e. if the input file name taken by $argv[1] is xyz.gz, I want
> > $argv[1] to be assigned as xyz.
>
> > How can I do this using cut command.
>
> > Thanks in advance
> > Unix Baby
>
> You could use the basename command like this:
> # basename a.gz .gz
> a
>
> Regards,
> --
> Medernach

Thanku Medernach. Perfect Solution!!!
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