Display mysql data

put your code and mention clearly what exactly you want

Hi!

I believe you want something like this:

<?php

// Not Tested!!!

if(isset($_GET['sites'])) {

	$id = (int)$_GET['sites'];
	
	if($id > 0) {

		$sql_about = sprintf("SELECT `FileName`, `About` FROM `sites` WHERE `ID` = %d LIMIT 1;", $id);
    	$qry_about = mysql_query($sql) or die (mysql_error());
    	
    	if(mysql_num_rows($qry_about) == 1) {

			$obj_about = mysql_fetch_object($qry_about);
			
			?>
			
			<fieldset>
				<legend>About <?php echo $obj_about->FileName; ?></legend>
				<p><?php echo $obj_about->About; ?></p>
			</fieldset>
			
			<?
		  mysql_free_result($obj_about);
    	}


	}
	
}


$sql = "SELECT `ID`, `FileName` FROM `sites` ORDER BY `FileName`;";
$qry = mysql_query($sql);

if(mysql_num_rows($qry) > 0) {

	while($obj = mysql_fetch_object($qry)) {

?>

		 <p> <a href="<?php echo $_SERVER['PHP_SELF']; ?>?sites=<?php echo $obj->ID; ?>"><?php echo $obj->FileName; ?></a> </p>

<?php

	}

  mysql_free_result($qry);
}

?>

Or perhaps this:

<?php

// Not Tested!!!

$sql = "SELECT `ID`, `FileName`, `About` FROM `sites` ORDER BY `FileName`;";
$qry = mysql_query($sql);

if(mysql_num_rows($qry) > 0) {

$id = 0;

 if(isset($_GET['sites'])) {

 	$id = (int)$_GET['sites'];

 }

	while($obj = mysql_fetch_object($qry)) {

		if($id > 0) { ?>

			<p> <a href="<?php echo $_SERVER['PHP_SELF']; ?>?sites=<?php echo $obj->ID; ?>"><?php echo $obj->FileName; ?></a> </p>

		<?php } else { ?>

			<p>
 			  <?php echo $obj->FileName; ?></a> <br />
 			  <strong>About:</strong> <?php echo $obj->About; ?>
			</p>

		<?php }

	}

  mysql_free_result($qry);
}

?>

I made it in two different ways, since I was not sure about how you want to do it.

I believe that it would had better presentation if you use JavaScript and make use of display style property to show the details and hidden them, anyway, hope PHP snippets would help you.