Regular expression problem (PERL Syntax)

Parse list and convert to HTML format

[Garfield]

Hi,

I'm having a tough problem that I want to solve with regular expressions. I'm pretty sure that it is possible for a regex to parse this thing, but since I only know the basic stuff, Im pretty much lost here.

I want to parse a string like this:

##
#list item1
#list item2
#list item3
##

And this is what I want the regex to return:

<ul>
<li>list item1</li>
<li>list item2</li>
<li>list item3</li>
</ul>

Basically, I want to parse my listing and return a valid HTML unordered list. The regex has to find start end end tags (##) and then parse the inner part to find strings like #text\n where \n is a line break.

Can anyone help me with this?

Thanks in advance!

[ironik]

This function will work for one list per text call:

function convert_to_list($text)
{
$pattern1 = "/(\#\#\n)((\#{1})[\w?\s?\W?\S?]+\n)*(\#\#\n)/";
$replace1 = "<ul>$2</ul>";

$list = preg_replace($pattern1, $replace1, $text);

$pattern2 = "/(\#{1})(\w{1,}.+)?\n/";
$replace2 = "<li>\\2</li>";

$text = preg_replace($pattern2, $replace2, $list);
return $text;
}

I haven't been able to get it working for multiple lists in the same piece of text... perhaps someone here is able to tweak the regex a little so it will do multiple replacements successfully?

I'm still discovering regex's myself, so it's probably not that efficient. :(

[Garfield]

Thank you!

I don't understand the regex completely, but I know what it does ;)

It would be great if anyone could help getting this regex work with multiple lists in a text string.

Thanks!

[ironik]

If you want to learn some more about regex's heres a couple of links. Handy to know, especially when filtering and validating content:

Sample regex's (regex library)
[regexlib.com...]

In depth Regex tutorial
[regular-expressions.info...]

O'reilly regex pocket reference (PDF)
[oreilly.com...]

More regex references (this is probably the best one for beginners)
[phpbuilder.com...]

[jatar_k]

Welcome to WebmasterWorld ironik

[ironik]

Thanks ;)

Been lurking for a while, thought it was time to sign up!

[andrew_m]

You made me curious if that's possible as a single regex. This seems to work:

#!/usr/bin/perl -w 
$a=<<EOT; 
## 
#list item1 
#list item2 
#list item3 
## 
another list 
## 
#another 1 
#another 2 
## 
EOT 
$a=~s{(^¦\r¦\n) 
  \#\#([\s\r\n]+) 
  ((?:\#.*?[\r\n])+?) 
  \#\#([\r\n\s]+) 
  } 
  { local $prefix=$1 . '<UL>' . $2; 
  local $suffix='</UL>' . $4; 
  (local $lines=$3)=~s/^#/<LI>/mg; 
  $prefix . $lines . $suffix; 
  }gsex; 
print $a; 

As a bonus you get 'gsex' (great sex) as a regex qualifier :)

[Garfield]

@andrew_m: Thanks for your help! I honestly have no idea what your regex is doing for most of the time.

I tried to run it as a PERL script, but it didnt work. The error was: Can't find string terminator "EOT" anywhere before EOF at ./perltest line 2.

How can I get this working inside PHP's preg_replace() function?

Thanks!

[andrew_m]

I have no idea, sorry, don't know a thing about php. I thought you asked about perl and it does work fine in perl -- it's cut-n-paste from the actual script that I ran to test it.