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Subject Author Date
pattern match Venkatesh can....can... 03-20-2008
---> Re: pattern match Gunnar Hjalmars...03-20-2008
---> Re: pattern match venkatesh.naughty03-20-2008
---> Re: pattern match Gunnar Hjalmars...03-20-2008
`--> Re: pattern match Tad J McClellan03-20-2008
Posted by Venkatesh can....can... on March 20, 2008, 1:29 am
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 $var="{' venkat'}->->";
i want to get the "yes" token;
if i use
$var=~/\{'( .* )\}$/
i get venkat'}->->{'yes
how to get the "yes" token...

Posted by Gunnar Hjalmarsson on March 20, 2008, 2:02 am
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 Venkatesh can....can... wrote:
> $var="{' venkat'}->->";
> i want to get the "yes" token;
> if i use
> $var=~/\{'( .* )\}$/
> i get venkat'}->->{'yes

No you don't. You get nothing, because that regex does not match.
However, with the /x modifier it matches and assigns the string you
mention to $1.

> how to get the "yes" token...

One way:

$var =~ /.+$/;

--
Gunnar Hjalmarsson
Email: http://www.gunnar.cc/cgi-bin/contact.pl

Posted by venkatesh.naughty on March 20, 2008, 2:18 am
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> Venkatesh can....can... wrote:
> > $var=3D"{' venkat'}->->";
> > i want to get the "yes" token;
> > if i use
> > $var=3D~/\{'( .* )\}$/
> > i get venkat'}->->{'yes
>
> No you don't. You get nothing, because that regex does not match.
> However, with the /x modifier it matches and assigns the string you
> mention to $1.
>
> > how to get the "yes" token...
>
> One way:
>
> =A0 =A0 =A0$var =3D~ /.+$/;
>
> --
> Gunnar Hjalmarsson
> Email:http://www.gunnar.cc/cgi-bin/contact.pl

@gunnar

thanks it works but how?
the first .+ is greedy know? it 'll match up to the end right?

Posted by Gunnar Hjalmarsson on March 20, 2008, 6:13 am
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venkatesh.naughty@gmail.com wrote:
>> Venkatesh can....can... wrote:
>>>
>>> $var="{' venkat'}->->";
>>> i want to get the "yes" token;
>>> if i use
>>> $var=~/\{'( .* )\}$/
>>> i get venkat'}->->{'yes
>>
>> No you don't. You get nothing, because that regex does not match.
>> However, with the /x modifier it matches and assigns the string you
>> mention to $1.
>>
>>> how to get the "yes" token...
>>
>> One way:
>>
>> $var =~ /.+$/;
>
> thanks it works but how?
> the first .+ is greedy know?

Yes.

> it 'll match up to the end right?

No, it matches one or more characters as long as it can without
preventing the whole regex from matching; in this case up to and
including the second arrow.

Remember that greediness never affects whether a regex matches or not.
It just may affect _how_ the regex matches.

--
Gunnar Hjalmarsson
Email: http://www.gunnar.cc/cgi-bin/contact.pl

Posted by Frank Seitz on March 20, 2008, 6:42 am
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Gunnar Hjalmarsson wrote:
> venkatesh.naughty@gmail.com wrote:
>>
>>the first .+ is greedy know?
>
> Yes.
>
>>it 'll match up to the end right?
>
> No, it matches one or more characters as long as it can without
> preventing the whole regex from matching; in this case up to and
> including the second arrow.

Internally .+ matches the whole string at first.
Then, the regex-engine backtracks until it achieves an
overall match (or not).

Frank
--
Dipl.-Inform. Frank Seitz; http://www.fseitz.de/
Anwendungen für Ihr Internet und Intranet
Tel: 04103/180301; Fax: -02; Industriestr. 31, 22880 Wedel

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