From: Robert Klemme <shortcutter@...>
Date: 2007-02-06T07:10:13+09:00
Subject: Re: Regular expressions (extracting urls)

On 05.02.2007 21:37, David Krmpotic wrote:
> Hi!
> 
> I have to extract an url from the text and make it a link (a href...)..
> The trick is that I have to be careful not to replace the url, that are
> already a part of the link.
> 
> so:
> 
> link = "Go here: http://www.something.com!"
> link.gsub!(/https?:\/\/[a-z0-9\.\-\_=&\+\/\?]+/i, '<a
> href=\'\0\'>\0</a>')
> 
> link becomes:
> link = "Go here: <a
> href='http://www.something.com'>http://www.something.com</a>!"
> 
> Now...
> 
> When the link is this:
> link = "Go here: <a
> href='http://www.something.com'>http://www.something.com</a>!"
> 
> The regular expression must not replace it!
> 
> I know that for example if I want to exclude the links that start with
> xhttp, I can write:
> link.gsub!(/([^x]https?:\/\/[a-z0-9\.\-\_=&\+\/\?]+)/i, '<a
> href=\'\1\'>\1</a>')
> 
> but how can I exclude links that start with href=" and href=' ?
> The problem is that I don't know how to specify that HREF cannot
> preceede the link (cannot write [^href], [^(href)] doesn't seem to work
> either and it also screws \n ... )

Maybe it's enough to do

s.gsub( /([^"'])\b(http[^\s"']+)\b([^"'])/,
   '\\1<a href="\\2">\\2</a>\\3' )

That depends on your input text.  This piece has some weaknesses, e.g. 
won't substitute hrefs at the beginning and end of the string (you could 
pad with a whitespace).

Kind regards

	robert