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Editorial Team
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Asked: 2026-05-20T20:54:21+00:00

I am a newbie to mootools and web development also. I have read this

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I am a newbie to mootools and web development also. I have read this pretty cool blog
and I want to extend the code to connect with a database to update the rating with a php file. but unfortunately my code not working means database is not updating. Can someone please explain me why. Thanks a lot…

Here’s the code

star.html

<html>

<script src="mootools-1.3.js"></script> 
<script src="lorenzos-MooStarRating-422072a/Source/moostarrating.js"></script> 
<script> 

    // Configure the image paths
    var MooStarRatingImages = {
        defaultImageFolder: 'lorenzos-MooStarRating-422072a/Graphics/',
        defaultImageEmpty:  'star_empty.png',
        defaultImageFull:   'star_full.png',
        defaultImageHover:  "star_boxed_hover.png"
    };

    // Post iD
    var postId = 10;

    // When the DOM is ready....
    window.addEvent("domready",function() {

        // Create our instance
        // Advanced options
        var advancedRating = new MooStarRating({
            form: 'ratingsForm',
            radios: 'rating',
            half: false,
            //imageEmpty: 'star_boxed_empty.png',
            //imageFull:  'star_boxed_full.png',
            //imageHover: "star_boxed_hover.png", 
            width: 17, 
            tip: 'Rate <i>[VALUE] / 7.0</i>', 
            tipTarget: $('htmlTip'),
            tipTargetType: 'html', 
            click: function(value) {
                // Send ajax request to server
                new Request.send({
                    url: "rateSave.php",
                    data: {'rating': value}
                });
            }
        });



    });

</script> 

<form name="ratingsForm">
<label>Select The Number of Stars</label>
<input type="radio" name="rating" value="1.0" checked="checked">

<input type="radio" name="rating" value="2.0">

<input type="radio" name="rating" value="3.0">

<input type="radio" name="rating" value="4.0">

<input type="radio" name="rating" value="5.0">

<input type="radio" name="rating" value="6.0">

<input type="radio" name="rating" value="7.0">

<!--<input type="radio" name="rating" value="7.5">
<input type="radio" name="rating" value="8.0">
<input type="radio" name="rating" value="8.5">
<input type="radio" name="rating" value="9.0">
<input type="radio" name="rating" value="9.5">
<input type="radio" name="rating" value="10.0">-->



<span id="htmlTip"></span>
</form>

</html>

rateSave.php

<?php

$con = mysql_connect("localhost","root","");
if (!$con){
  die('Could not connect: ' . mysql_error());
}
mysql_select_db("rating", $con);

$starCount =$_POST['rating'];

$result=mysql_query("INSERT INTO star VALUES('hotel','$starCount')");



mysql_close($con);

?>
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1 Answer

  1. Editorial Team

    Hi Pavithra Gunasekara, the error is ‘nothing’, here : 

    click: function(value) {
// Send ajax request to server ...
}

    instead of ‘click’, the name of the CallBack function is onClick i.e.

    onClick: function(value) {
// Send ajax request to server ...
}

    about the ‘click’, you could do this way i.e.

    advancedRating.addEvent('click', function(){ new Request.send({/* ... */}) });

    working example with ‘onClick’ instead of the ‘click’ inside the new instance definition: http://jsfiddle.net/steweb/LDw4y/

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