Help with preg_replace

Crussell

Hi all,

For reasons best known to myself I need to take a source file, read it into a PHP script, pick out all instances of a variable called $ST and wrap them in DIV tags, so imagine a file like:

<h1> Welcome to $ST['site'], have a nice visit you $ST['epithet']

I want to wrap DIV tags around $ST['site'], $ST['epithet'] and any other $ST['INSERTSTUFFHEREs'] that I find, including the bit between ' and ' as the name in the DIV, so the output will include:

Being hopeless with regex doesn't help and I've got a partial expression which matches the string but doesn't do the replace - can anyone help shed light on how I might actually sort this out before I lose the will to live?

TIA

Chris

remnant

The main problem here is that you can't search for $ST['variable'] because it only exists in the string as the varialbes value. That being said, this can be done with regex but it's not necessary.

Something like this should work:

foreach ($ST as $key => $value) {
    $divTag = "<div name='$key'>$value</div>";

$text = str_replace($value, $divTag, $text);
}

Bear in mind, however, that if your variable values are not unique to the text other text will be replaced. This would be true even of a regex solution barring some other identifying piece of information.

HTH

obsidian

actually, you can do what he's after since he states he'll be searching through the PHP source itself. so, with that being said, i'd recommend something like this:

$String = '<h1> Welcome to $ST[\'site\'], have a nice visit you $ST[\'epithet\']</h1>';
$find = '|\$ST\[\'(.+?)\'\]|';
$replace = '<div name="$1">$ST[\'$1\']</div>';
$String = preg_replace($find, $replace, $String);

just tested it, and it works with every sample i ran through

Crussell

Guys, many thanks, two solutions to one problem is fantastic (and probably shows I should have thought harder first 🙂.

Cheers

Chris