I know I can refer in replacement to dynamic parts of the term in regex in PHP:
preg_replace('/(test1)(test2)(test3)/',"$3$2$1",$string);
(Somehow like this, I don't know if this is correct, but its not what I am looking for)
I want that in the regex, like:
preg_match_all("~<(.*)>.*</$1>~",$string,$matches);
The first part between the "<" and ">" is dynamic (so every tag existing in html and even own xml tags can be found) and i want to refer on that again in the same regex-term.
But it doesn't work for me. Is this even possible? I have a server with PHP 5.3
/edit:
my final goal is this:
if have a html-page with e. g. following source-code: HTML
<html>
<head>
<title>Titel</title>
</head>
<body>
<div>
<p>
p-test<br />
br-test
</p>
<div>
<p>
div-p-test
</p>
</div>
</div>
</body>
</html>
And after processing it should look like
$htmlArr = array(
'html' => array(
'head' => array('title' => 'Titel'),
'body' => array(
'div0' => array(
'p0' => 'p-test<br />br-test',
'div1' => array(
'p1' => 'div-p-test'
)
)
)
));
Placeholders in the replacement string use the $1 syntax. In the regex itself they are called backreferences and follow the syntax \1 backslash and number.
http://www.regular-expressions.info/brackets.html
So in your case:
preg_match_all("~<(.*?)>.*?</\\1>~",$string,$matches);
The backslash is doubled here, because in PHP strings the backslash escapes itself. (In particular for double quoted strings, else it would become an ASCII symbol.)