preg_replace backreferencing problem

[Terence]

I've been puzzling over something for ages and now give up:

Does anyone see any problem with any of the following:

$line = preg_replace('/\[#([A-Fa-z0-9]{1,6})\]/i', "<font color='#
${1}'>", $line);

$line = preg_replace('/\[(#(([A-Fa-f0-9]){3}(([A-Fa-f0-9]){3})?))\]/
i', "<font color='#${1}'>" , $line);

I tried both ; both don't work.

This snippet should convert something like [#FF0000] (hex colours)
into an HTML font tag. I believe the reg.exp IS matching (since I do
get a <fonttag), but the backreferencing is not working since the
font tag would contain an empty hex code.

I've tried using \\1 for the backreferencing , or even \\0 just to test
- but even that doesn't work.

Is it possible that backreferencing can just be turned off?!?

- Terence

[Bocah Sableng]

On May 25, 6:24 am, Terence <terence.par... @gmail.comwrote :

I've been puzzling over something for ages and now give up:

Does anyone see any problem with any of the following:

$line = preg_replace('/\[#([A-Fa-z0-9]{1,6})\]/i', "<font color='#
${1}'>", $line);

try:
$line = preg_replace('/\[#([A-Fa-z0-9]{1,6})\]/i', "<font color='#
$1'>", $line);

[gosha bine]

On 25.05.2007 01:24 Terence wrote:

I've been puzzling over something for ages and now give up:

Does anyone see any problem with any of the following:

$line = preg_replace('/\[#([A-Fa-z0-9]{1,6})\]/i', "<font color='#
${1}'>", $line);

$line = preg_replace('/\[(#(([A-Fa-f0-9]){3}(([A-Fa-f0-9]){3})?))\]/
i', "<font color='#${1}'>" , $line);

I tried both ; both don't work.

This snippet should convert something like [#FF0000] (hex colours)
into an HTML font tag. I believe the reg.exp IS matching (since I do
get a <fonttag), but the backreferencing is not working since the
font tag would contain an empty hex code.

I've tried using \\1 for the backreferencing , or even \\0 just to test
- but even that doesn't work.

Is it possible that backreferencing can just be turned off?!?

- Terence

$line = preg_replace('/\[(#[A-F0-9]+)\]/i', "<font color='$1'>", $line);

the correct syntax is $1 not ${1} or similar.

--
gosha bine

extended php parser ~ http://code.google.com/p/pihipi
blok ~ http://www.tagarga.com/blok