On Jun 18, 5:54 pm, [EMAIL PROTECTED] (John Degen) wrote:
> >----- Original Message ----
> >From: Paul Lalli <[EMAIL PROTECTED]>
> >To: [EMAIL PROTECTED]
> >Sent: Monday, June 18, 2007 6:47:05 PM
> >Subject: Re: Command line usage
>
> >On Jun 18, 10:50 am, [EMAIL PROTECTED] (John Degen) wrote:
>
> >> I think I'm out of luck with this OS;) Your suggestion for creating a
> >> backup
> >> file gave the same result: no error, no change in the files. The output of
> >> 'perl -le"print for @ARGV" *' is * and the other is *.*. Funny though that
> >> sed *does* work.
>
> >Ah. Well there's your problem. The command line interpreter you're
> >running doesn't expand wildcards. That's why Perl wasn't giving you
> >any errors - it had nothing to do because there was no file named "*"
> >that it could find... You'll have to expand the wildcard from within
> >Perl.
>
> >perl -pi.bkp -e"BEGIN { @ARGV = glob($ARGV[0]); } s/old/new/;" *
>
> >Hope this helps,
> >Paul Lalli
>
> Thank you all for your responses. Paul's suggestion above is the winner. The
> command works, creates a neat backup file and does what it's told. I'll be
> checking the docs to see *why* it works:)
Normally your command line interpreter would expand "*" into a list of
files, and perl would store them in @ARGV, e.g.,
$ARGV[0] = "a.txt";
$ARGV[1] = "b.txt";
$ARGV[2] = "c.txt";
Instead, yours is not expanding, but simply passing "*" unchanged, but
perl still stores it in @ARGV, which looks like
$ARGV[0] = "*";
So in the BEGIN block (which executes before the -p flag starts
reading files), Paul's code passes $ARGV[0] to the glob() function,
which is perl's way of expanding "*" into a list of files. Assigning
that output to @ARGV (overwriting $ARGV[0] in the process) yields,
e.g.,
$ARGV[0] = "a.txt";
$ARGV[1] = "b.txt";
$ARGV[2] = "c.txt";
Q.E.F.
--
Brad
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