I may be off base here, but given the last post of yours I read, it sounds like you want us to do the work for you. You obviously know some PHP and the sample code you gave basically has what you want; when you get a request with querystring pageid=5 you just want to pull out a different field from the database and since you already know how to get info from the database, that's not the issue; in fact, I'm not really sure what the problem is.
please help me what should i do
following is is my menu.php code
<div id="menu">
<div id="menumain">
<?php
include_once('config.php');
$result = mysql_query("SELECT * FROM page ");
while($row = mysql_fetch_array($result))
{
?>
<a href="?pageid=<?php echo $row['id']; ?>"><?php echo $row['pagename']; ?> </a>
<?php
}
?>
</div>
</div>
following is my index.php code
<?php include_once('config.php');?>
<?php include_once('header.php'); ?>
<?php include_once('menu.php');?>
<div id="content">
<?php include_once('sidebar.php'); ?>
<div id="maincontent">
<!-- include_once('config.php');-->
<?php
if(isset($_GET['pageid']))
{
$result=mysql_query("select content from page where id='pageid'");
echo '$result';
}
?>
</div>
<div id="space"></div>
</div>
<?php include_once('footer.php'); ?>