On Jun 3, Balint, Jess said:
Hello all. I am working on my script, and I would like to be able to use
STDIN as a filehandle input unless a filename is specified. I have used the
getopt for the filename, let's say $opt_f.
I would use the magic of @ARGV and , and do this:
@ARGV = $filename if
On Monday, June 3, 2002, at 09:27 AM, Jeff 'japhy' Pinyan wrote:
On Jun 3, Balint, Jess said:
Hello all. I am working on my script, and I would like to be able to use
STDIN as a filehandle input unless a filename is specified. I have used
the
getopt for the filename, let's say $opt_f.
On Jun 3, bob ackerman said:
@ARGV = $filename if $opt_f;
why '' ? isn't '$filename' enough for get '' to open and read from that
file?
I feel safer when I explicitly state the mode. Unless you want the user
to put 'foo |' as the filename (that is, the output of a pipe), then you
should
On Monday, June 3, 2002, at 10:37 AM, Jeff 'japhy' Pinyan wrote:
On Jun 3, bob ackerman said:
@ARGV = $filename if $opt_f;
why '' ? isn't '$filename' enough for get '' to open and read from
that
file?
I feel safer when I explicitly state the mode. Unless you want the user
to put
[mailto:[EMAIL PROTECTED]]
Sent: Monday, June 03, 2002 1:38 PM
To: bob ackerman
Cc: [EMAIL PROTECTED]
Subject: Re: Conditional Operator STDIN
On Jun 3, bob ackerman said:
@ARGV = $filename if $opt_f;
why '' ? isn't '$filename' enough for get '' to open and
read from that
file
On Monday, June 3, 2002, at 10:45 AM, bob ackerman wrote:
On Monday, June 3, 2002, at 10:37 AM, Jeff 'japhy' Pinyan wrote:
On Jun 3, bob ackerman said:
@ARGV = $filename if $opt_f;
why '' ? isn't '$filename' enough for get '' to open and read from
that
file?
I feel safer when
