preg_replace只有OUTSIDE标签？ （...我们不是在谈论完整的'html解析'，只是一点降价）

Question

What is the easiest way of applying highlighting of some text excluding text within OCCASIONAL tags "<...>"?

CLARIFICATION: I want the existing tags PRESERVED!

$t = 
preg_replace(
  "/(markdown)/",
  "<strong>$1</strong>",
"This is essentially plain text apart from a few html tags generated with some
simplified markdown rules: <a href=markdown.html>[see here]</a>");

Which should display as:

"This is essentially plain text apart from a few html tags generated with some simplified markdown rules: see here"

... BUT NOT MESS UP the text inside the anchor tag (i.e. <a href=markdown.html> ).

I've heard the arguments of not parsing html with regular expressions, but here we're talking essentially about plain text except for minimal parsing of some markdown code.

Answer 1

Actually, this seems to work ok:

<?php
$item="markdown";
$t="This is essentially plain text apart from a few html tags generated 
with some simplified markdown rules: <a href=markdown.html>[see here]</a>";

//_____1. apply emphasis_____
$t = preg_replace("|($item)|","<strong>$1</strong>",$t);

// "This is essentially plain text apart from a few html tags generated 
// with some simplified <strong>markdown</strong> rules: <a href=
// <strong>markdown</strong>.html>[see here]</a>"

//_____2. remove emphasis if WITHIN opening and closing tag____
$t = preg_replace("|(<[^>]+?)(<strong>($item)</strong>)([^<]+?>)|","$1$3$4",$t);

// this preserves the text before ($1), after ($4) 
// and inside <strong>..</strong> ($2), but without the tags ($3)

// "This is essentially plain text apart from a few html tags generated
// with some simplified <strong>markdown</strong> rules: <a href=markdown.html>
// [see here]</a>"

?>

A string like $item="odd|string" would cause some problems, but I won't be using that kind of string anyway... (probably needs htmlentities(...) or the like...)