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PHP Automated Tooltips with MySQL

Hi everyone I hope you are all well.

Guys i have the following setup.

1 MySQL Table called: tooltip_ttp
------------------------------------------------------------
It has 3 fields, with the following rows currently..
1) id_ttp (autonum) 2) name_ttp (varchar) 3) description_ttp (text)
rows: 1 Ajax This is Ajax
2 HTML This is HTML
3 CSS This is CSS
4 PHP This is PHP
etc

I have 1 html page: ajax-tooltip.html
------------------------------------------------------------
This calls a javascript function on mouseover of a link, which displays the description_ttp field data from the row given in the query string, for example:

<a href="#" onmouseover="ajax_showTooltip('demo-pages/ajax-tooltip.php?id=4',this);return false" onmouseout="ajax_hideTooltip()">PHP</a></td>

1 php page called ajax-tooltip.php
------------------------------------------------------------
This gets the posted value of 'id' in the query string from ajax-tooltip.html, then uses it to run a mysql query to return the right row and its description to display in a tooltip on the page ajax-tooltip.html

##############################
WHAT IM TRYING TO DO:
##############################

Currently, in ajax-tooltip.html, im hard coding a hrefs along with a hard coded query string id number for every link that i want to display a tooltip for from the mysql table. I dont want to have to hard code query string values nor have to duplicate javascript on mouseover code

What i would like is the following...

In ajax-tooltip.html,

Display in a table (or div) a list of every record (the name_ttp field) as a link.
On mouseover of each of those links, the corresponding description_ttp text is displayed in the tooltip.

So basically,
currently...

<a href="#" onmouseover="ajax_showTooltip('demo-pages/ajax-tooltip.php?id=4',this);return false" onmouseout="ajax_hideTooltip()">PHP</a></td>

going to what id like which would be something like:

<link (name_ttp field ) > on mouseover, display description_ttp for this row in a tooltip
<link (name_ttp field ) > on mouseover, display description_ttp for this row in a tooltip
<link (name_ttp field ) > on mouseover, display description_ttp for this row in a tooltip
<link (name_ttp field ) > on mouseover, display description_ttp for this row in a tooltip
<link (name_ttp field ) > on mouseover, display description_ttp for this row in a tooltip
etc

Each of the above lines eg. the links with name_ttp field, and the onmouseover, ideally would be automatically created for every record from the mysql table, along with its associated query string id value.

Any help greatly appreciated.

Guys i have attached the code in the snippet window for the files.

Thank you.

==================================== ajax-tooltip.html
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">
<html>
<head>
	<title>Ajax tooltip</title>
	<script type="text/javascript" src="js/ajax-dynamic-content.js"></script>
	<script type="text/javascript" src="js/ajax.js"></script>
	<script type="text/javascript" src="js/ajax-tooltip.js">	
	</script>	
	<link rel="stylesheet" href="css/ajax-tooltip.css" media="screen" type="text/css">
	<link rel="stylesheet" href="css/ajax-tooltip-demo.css" media="screen" type="text/css">
</head>
<body>
 
<div id="mainContainer">
	<div id="menuColumn">		
	</div>
	<div id="mainContent">
		<table width="100%" class="productTable">
			<thead>
				<th>Product</th>
				<th>Info</th>
			</thead>
			<tbody>
				<tr class="oddRow">
					<td>JS Calendar</td>
					<td>
						<a href="#" onmouseover="ajax_showTooltip('demo-pages/ajax-tooltip.php?id=4',this);return false" onmouseout="ajax_hideTooltip()">PHP</a></td>	
				</tr>			
				</tr>
			</tbody>
		</table>	
		<div id="debug"></div>
	</div>
</div>
 
</body>
</html>
 
============================================ ajax-tooltip.php
<?php
$dbServer="localhost";
$dbName="foo";
$link = mysql_connect( $dbServer, 'root', 'password');
if (!$link) {
    die('Not connected : ' . mysql_error());
}
 
// make foo the current db
$db_selected = mysql_select_db($dbName, $link);
if (!$db_selected) {
    die ("Can't use $dbName : " . mysql_error());
}
 
$result = mysql_query("SELECT description_ttp FROM tooltip_ttp WHERE id_ttp=" . $_GET['id'] ) or die( mysql_error());
 
$row=mysql_fetch_assoc($result);
echo $row['description_ttp'];
mysql_close();
?>

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babuno5

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Simon336697

ASKER

Hi babuno5,
Thanks so much mate...that is terrific thank you.