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From: George on 5 Feb 2010 21:56
Hi
I am trying to get directory name from say a path
c:/app/organizer
what I want to get is organizer , so basically if I search from end
of string and then lookbehind
($name)=($path=~(/(?<=\/)(.*?)(.*?)\z/));

but it does not result in getting to oraginzer , it fetches app/
organizer , kind of greedy

I know I can do it using FILE::BASENAME , but want to understand how
to do it using regexp and look back
thaks
/g
From: Tad McClellan on 5 Feb 2010 23:08
George <iamhello99(a)gmail.com> wrote:

> I am trying to get directory name from say a path
> c:/app/organizer
> what I want to get is organizer ,

(my $name = $path) =~ s!.*/!!;
or
my($name) = $path =~ m!([^/]+)$!;


> so basically if I search from end
> of string and then lookbehind
> ($name)=($path=~(/(?<=\/)(.*?)(.*?)\z/));
>
> but it does not result in getting to oraginzer , it fetches app/
> organizer ,


No it doesn't.


> I know I can do it using FILE::BASENAME , but want to understand how
> to do it using regexp and look back


I don't see that lookbehind is necessary.


--
Tad McClellan
email: perl -le "print scalar reverse qq/moc.liamg\100cm.j.dat/"
From: Xho Jingleheimerschmidt on 5 Feb 2010 23:58
George wrote:
> Hi
> I am trying to get directory name from say a path
> c:/app/organizer
> what I want to get is organizer , so basically if I search from end
> of string and then lookbehind
> ($name)=($path=~(/(?<=\/)(.*?)(.*?)\z/));

How would the zero-width look behind change anything? If the slash
itself isn't captured in $1 or $2, who cares if it is in $& ?

>
> but it does not result in getting to oraginzer , it fetches app/
> organizer , kind of greedy

In my hands, $name ends up as the empty string. What you indicate ends
up in $2, not in $1 or $name.

> I know I can do it using FILE::BASENAME , but want to understand how
> to do it using regexp and look back
> thaks
> /g

/([^\/]*)\z/

No look behind needed.

Xho
From: C.DeRykus on 6 Feb 2010 08:09
On Feb 5, 6:56 pm, George <iamhell...(a)gmail.com> wrote:
> Hi
> I am trying to get directory name from say a path
> c:/app/organizer
> what I want to get is organizer , so basically  if I search from end
> of string and then lookbehind
> ($name)=($path=~(/(?<=\/)(.*?)(.*?)\z/));
>
> but it does not result in getting to oraginzer , it fetches app/
> organizer , kind of greedy
>
> I know I can do it using FILE::BASENAME , but want to understand how
> to do it using regexp and look back

Here's a possible lookbehind solution:

( $name ) = $path =~ m{ (?<=/) ([^/]*) \z }x;


or, with 5.10 you could use \K (perldoc perlre):

( $name ) = $path =~ m{ / \K ([^/]*) \z }x;



But, the above are somewhat obfuscated because there's a
straightforward, simpler alternative:

( $name ) = $path =~ m{ / ([^/]*) \z }x


The only difference is that $& now includes '/'; whereas,
the latter doesn't.

--
Charles DeRykus
From: Dr.Ruud on 6 Feb 2010 08:56
C.DeRykus wrote:

> ( $name ) = $path =~ m{ / ([^/]*) \z }x

That $name will end up false if there is no slash in $path.

perl -Mstrict -le '
for my $path (undef, "", "test", "test/123") {
my ($name) = $path =~ m~([^/]+)\z~;
print defined($name) ? "<$name>" : "undef";
}
'
undef
undef
<test>
<123>

But you should just use File::Basename for this.
(beware, the module name was grossly misspelled by OP)

--
Ruud
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