Le 09/01/2020 à 17:26, Derek Jones a écrit :
To whom it may concern,
Please remove this email address (Derek Jones) from the modperl email
list.
Thanks!
Hi,
have a look at:
http://perl.apache.org/maillist/modperl.html#toc_Subscription_Information
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To whom it may concern,
Please remove this email address (Derek Jones) from the modperl email list.
Thanks!
It probably won't ever work. The problem is the * (star/asterisk) after
\1 is being interpreted in the context of regular expressions, ie,
"zero-or-more matches", and not as a multiplication operator. In fact,
~]$ perl -Mstrict -le 'my $str = "2 3 23"; print "true" if
I don't know which database you're using but we use Apache::DBI + PGBouncer
On Thu, Jan 9, 2020 at 1:08 AM Mithun Bhattacharya wrote:
> Connection pooling is implemented in DBI and enabled on need basis and in
> Apache::DBI it is enabled by default.
>
> On Wed, Jan 8, 2020 at 9:04 PM Wesley
On Thu, Jan 09, 2020 at 05:21:38PM +0800, Wesley Peng wrote:
> what does (??{$1*$2}) means?
Check the perlre(1) man page for explanation of "(??{ code })".
As already mentioned, other venues like Perlmonks might serve better
for the generic Perl questions.
--
Jan Pazdziora
what does (??{$1*$2}) means?
Thanks.
on 2020/1/9 12:49, Joseph He wrote:
I think $str =~ /(\d+)\s(\d+)\s(??{$1*$2})/ should do it
My Perl version is v5.26.1
My apologies you were right. I misunderstood the question. Yves
On Thu, 9 Jan 2020, 09:39 demerphq, wrote:
> On Thu, 9 Jan 2020 at 05:49, Joseph He wrote:
> >
> > I think $str =~ /(\d+)\s(\d+)\s(??{$1*$2})/ should do it
> > My Perl version is v5.26.1
>
> I think you mean (??{ "$1*$2"})
Oh. Heh. I misunderstood the intent. I thought you wanted to match a
sequence of digits followed by a space followed by more digits followed by
a space followed by the first set of digits repeated 0 or more times
followed by the second set of digits. If you want multiplication then
Joseph He's
Hallo
on 2020/1/9 16:35, demerphq wrote:
$str=~/(\d+)\s(\d+)\s(\1*\2)/
$1 refers to the capture buffers from the last completed match, \1
inside of the pattern part of a regex refers to the capture buffer of
the currently matching regex.
This doesn't work too.
perl -Mstrict -le 'my $str =
On Thu, 9 Jan 2020 at 05:49, Joseph He wrote:
>
> I think $str =~ /(\d+)\s(\d+)\s(??{$1*$2})/ should do it
> My Perl version is v5.26.1
I think you mean (??{ "$1*$2"}) which might work, but it will be error
prone, (??{"(?$1)*$2") would be better, but both will be slow, as each
time a new
This isnt really the forum for random perl help, i suggest Perlmonks instead.
$str=~/(\d+)\s(\d+)\s(\1*\2)/
$1 refers to the capture buffers from the last completed match, \1
inside of the pattern part of a regex refers to the capture buffer of
the currently matching regex.
Yves
On Thu, 9 Jan
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