Re: Executebale code from a databse

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 contain html and then we reopen <?php again) So if you have a file: <?php include('test.php'); ?> 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" <newsaddress@saracenvsu.org.uk> schrieb im Newsbeitrag news:20020522092539.43795.qmail@pb1.pair.com... > 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 I > 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 which > version! (fairly recent though) > >