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PHP Automated Tooltips with MySQL

Posted on 2008-10-09
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Last Modified: 2012-05-05
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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Comment
Question by:Simon336697
2 Comments
 
LVL 15

Accepted Solution

by:
babuno5 earned 500 total points
ID: 22685327
you will have to re write ajax-tooltip.html as  ajax-tooltip_display.php

sample code for ajax-tooltip_display.php is shown below


<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>

										<table border =1>

				

										<?

											$result = mysql_query("SELECT id_ttp,name_ttp FROM tooltip_ttp" ) or die( mysql_error());

											 

											while ($row = mysql_fetch_assoc($result)) {

												echo "<tr><td>";

												echo "<a href=\"#\"  onmouseover=\"ajax_showTooltip('demo-pages/ajax-tooltip.php?id=".$row["id_ttp"]."',this);return false\" onmouseout=\"ajax_hideTooltip()\">".$row["name_ttp"]."</a>";

												echo "</td></tr>";
 

											}
 

										?></table>		

										</td>   

                                </tr>                   

                                </tr>

                        </tbody>

                </table>        

                <div id="debug"></div>

        </div>

</div>

 

</body>

</html>

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LVL 1

Author Comment

by:Simon336697
ID: 22685387
Hi babuno5,
Thanks so much mate...that is terrific thank you.
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