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]

Re: Regex, how do I replace quotation pairs into &lt;LI&gt; &amp; &lt;/LI&gt;?

Kelvin wrote:[color=blue]
> Basically, my texts consists of normal text stream and some
> quotations.[/color]

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]

Re: Regex, how do I replace quotation pairs into &lt;LI&gt; &amp; &lt;/LI&gt;?

On Thu, 21 Oct 2004 18:27:05 +0800, Kelvin wrote:[color=blue]
> s/\".*?\"/<li>.*?<\/li>/g;[/color]

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 <toreau@gmail.c om>
"Time only seems to matter when it's running out." (Peter Strup)

[Gerhard M]

Re: Regex, how do I replace quotation pairs into &lt;LI&gt; &amp; &lt;/LI&gt;?

"Kelvin" <thefatcat28@ho tmail.com> wrote in message news:<41778d70$ 1@news.starhub. net.sg>...[color=blue]
> Tried ___s/\".*?\"/<li>.*?<\/li>/g;___ but not working.[/color]

hi kevin

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

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

gerhard