RE: Shell question

[jmmckee <jmmckee@xxxxxxxxxxxxxx>]

Very shortly after I posted my question, an answer came to me.  It is not elegant, but it works.  Actually a bit ugly.  The sed command is nicer.  What works is this:

        file='20'`ls $i|cut -b 5,6``ls $i|cut -b 1-4``ls $i|cut -b 7-10`

Thanks

John McKee


-----Original message-----
From: "Dennis Lovelady" iseries@xxxxxxxxxxxx
Date: Thu, 19 Nov 2009 11:39:31 -0600
To: "'Midrange Systems Technical Discussion'" midrange-l@xxxxxxxxxxxx
Subject: RE: Shell question

> > Given a file named like this:  MMDDYY.TXT I want to rename (or copy it 
> > ) to 20YYMMDD.TXT
> >
> > Might be multiple files, so I would do this:
> >
> > for i in *.TXT
> >     do
> >         cp $i '20'...
> >     done
>
>
> There are a few ways to do this.  One is directly in the shell, using
> something like ${varname#.TXT} but I'm not sure the # character is correct
> (could be % or !, maybe?) because I don't have my Korn reference handy.  You
> can find that online at kornshell.com.
>
> The next would be to pipe the filename through sed:  $workname=$(echo
> ${filename}|sed 's/.TXT//')
>
> There are many other ways to accomplish the same thing, but basically what
> you want to do is to get a list of the files, then "for each item" get the
> workname, and work at the workname level.
>
> Dennis Lovelady
> http://www.linkedin.com/in/dennislovelady
> --
> My intuition nearly makes up for my lack of good judgment.
>
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