eval ('?>'.$var.'<?php');
if you want to eval usual php scripts.
(We close the ?> then comes the content of the php script which also can
 html and then we reopen <?php again)

So if you have a file:

and you say
$var = "<?php include('test.php'); ?>";

you'll result in

eval("?><?php include('test.php'); ?><?php");

which will evaluate normally.

Regards Michael

"Peter" <[EMAIL PROTECTED]> schrieb im Newsbeitrag
> Hi.
> I'm changing my website to one based on My-SQL which will help with
> organization and searching etc. Hopefully, the code for all the pages will
> be stored in the database too.
> However, I cannot get PHP to parse / execute the code stored in the
> database. The script
> $query = mysql_query("SELECT * FROM pages", $link);
> $result = mysql_fetch_array($query);
> print $result['4'];
> gets the content of the page (column 4 of the database) but displays
> include("common/counter.php"); include("common/navbar.php");
> to the screen instead of opening and including these two files in the
> output.
> Is there something I need to do to the result to make it executable? Might
> need a \n between the two lines of code?
> I'm using Win 98, Apache 1.3.19, PHP 4.2.0 and MySQL but I'm not sure
> version! (fairly recent though)

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