I was trying to understand what "[^/]*" did in your regex. I know that you said that it was to make sure that there was not any "/" after the 7 characters. However I could not see the mechanics on how it would work.
Basically, I wanted a regex which would simply extract a pathname from a fully qualified filename and determine if "[^/]*" was needed to obtain the complete pathname.
So I modified your regex to extract the pathname and throw away the basename ... this showed me that
"[^/]*" was not required to traverse up to the last "/". The " .*/" part of the regex is greedy and grabs everything up to and including the last "/". This meant that "[^/]*" could be replaced with ".*" in your regex and still do the same thing...
----- Original Message ----- From: "Chris Devers" <[EMAIL PROTECTED]>
To: "Bernard Kenik" <[EMAIL PROTECTED]>
Cc: "Perl Beginners List" <[EMAIL PROTECTED]>
Sent: Monday, October 25, 2004 12:30 AM
Subject: RE: how the print the first 7 letter of file name
On Sun, 24 Oct 2004, Bernard Kenik wrote:
The part that puzzled me was [^/] .. so I experimented a little. I asked myself how I would extract the path portion.
Wasn't the requirement to *throw away* the path portion, not save it?
The final answer I came up with was: ( my $path = $file ) =~ s#(.*/).*#$1#;
This extract all characters up to and including the last slash since this is a
greedy regex.
Wouldn't that save the path, but not the filename?
Isn't that the opposite of what the guy was asking for?
Therefore I concluded that "[^/]" would not be needed.
( my $name = $file ) =~ s#.*/(.......).*$#$1#; does the same thing as ( my
$name = $file ) =~ s#.*/(.......)[^/]*$#$1#;
Maybe so, but when I was testing it on the command line, the variant you show was matching the first slash and the first 7 characters after it, which isn't what the guy (seemed to be) asking for.
That said, I'm still not sure I had it right. He said he wanted the first seven characters, which is what I was trying to write, but it looks like he wanted the part of the filename before the underscore:
$no_path_file = "NewProcess_date_22-oct-2004.log"; ( my $prefix = $no_path_file ) =~ s/_.*//; print $prefix; print $prefix;
-- prints "NewProcess", which seems to be what he really wanted.
But he *still* never described the goal clearly, so that's as close as I care to bother getting :-)
This is not meant as a criticism but simply to show that someone was watching and learning.
We're all learning, hopefully -- I know I am... :-)
--
Chris Devers
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