[Resolved] parse without output

[Herk]

I am writing a dynamic site engine sort of file. To be brief, I have part of the websites html stored in a db. In this html I have written to allow

[menu]

then I preg_replace that with the dynamic html for the menu.

Here is the code. The file it includes is the menu html with php variables.

/****************************************
|      Getting Template Variables       |
****************************************/
$stylevariable = @mysql_query("SELECT * FROM template WHERE site = '".$sitename."'");
$style = mysql_fetch_array($stylevariable);
// Template variables are now $style[nameindatabase]
$style['header'] = preg_replace("/\[date\]/", date('d M y, Hi T'), $style['header']);
$menu = include($htmlpath."/resources/menupre.php");
$style['header'] = preg_replace("/\[menu\]/", $menu, $style['header']);

Here is the problem. Since the $menu var that [menu] is replaced with is defined by including php with html it sends the headers.

How can I keep this html from being output until I call for it?

Herk

Reading through that, it sounds even to me like gibberish, but I don't know a better way to explain what I mean. If you understand please help.

[tuf-web_co_uk]

php's output buffering will do the job here

i havent had much experience with it but enough to know what lines you need to be on

<?

ob_start();
// Enter here php code you want to execute

// example
$menu = load_menu();
echo $menu;
$ob_contents = ob_get_contents();

ob_end_clean();

?>

for more information http://uk2.php.net/manual/en/ref.outcontrol.php

[Herk]

Outstanding!!!

Thank you kindly. I hate it when I get hung up like that. Glad there is a website with much more experienced people than I.

Herk