Regex, how do I replace quotation pairs into <LI> & </LI>?

[Kelvin]

Basically, my texts consists of normal text stream and some quotations.

This is my text stream, and inside "this streams" there are some "quotation
pairs"
which are to be replaced like this: <LI>this streams</LI> for formatting in
HTML.

Tried ___s/\".*?\"/<li>.*?<\/li>/g;___ but not working.

Thanks.
Kelvin

[Gunnar Hjalmarsson]

Kelvin wrote:

Basically, my texts consists of normal text stream and some
quotations.

Answered in comp.lang.perl.misc only, since this group is defunct.

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

[Tore Aursand]

On Thu, 21 Oct 2004 18:27:05 +0800, Kelvin wrote:

s/\".*?\"/<li>.*?<\/li>/g;

No need to escape those "-characters, AFAIK. And you don't want to
replace .*? above with - uhm - the regular expression .*?, do you?

Untested, but I think something like this should do it;

s,"(.*?)",<li>$1</li>,g;

Please read these:

perldoc perlretut
perldoc perlre
--
Tore Aursand <to****@gmail.com>
"Time only seems to matter when it's running out." (Peter Strup)

[Gerhard M]

"Kelvin" <th*********@hotmail.com> wrote in message news:<41********@news.starhub.net.sg>...

Tried ___s/\".*?\"/<li>.*?<\/li>/g;___ but not working.

hi kevin

try
s#"([^"]*)"#<li>$1</$1>#g

matches " (any text but quotes) "
and places (any text..) between <li> and </li>

gerhard