Help With JQuery Star Rating.

Jack Franklin · #1 (**permalink**) Feb 18th, 2008, 15:10

I'm trying to get the star query work in my PHP page.

<?php
include('admin/database.php'); ?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>The Best Flash Gallery</title>
<script src="jquery-latest.js" type="text/javascript" language="javascript"></script>
<script src="jquery.rating.js" type="text/javascript" language="javascript"></script>
<link type="text/css" rel="stylesheet" href="jquery.rating.css"/>
</head>

Am I missing a javascript src or something?

And this is bit with the form and the javascript.

Code: Select all

<body>
<script type="text/javascript">$(function(){ $('input[@type=radio].star').rating(); });</script>
<script>
$(function(){
 $('form').submit(function(){
  $('.test',this).html('');
  $('input',this).each(function(){ if(this.name!='') $('.test',this.form).append(''+this.name+': '+this.value+'<br/>'); });
  return false;
 });
});
</script>
<form method="post" action="">
<input class="star" type="radio" name="test-1-rating-2" value="1"/>
    <input class="star" type="radio" name="test-1-rating-2" value="2"/>
    <input class="star" type="radio" name="test-1-rating-2" value="3"/>
    <input class="star" type="radio" name="test-1-rating-2" value="4"/>
    <input class="star" type="radio" name="test-1-rating-2" value="5"/>
</form>
 
</body>
</html>

Is that all ok? I didn't quite understand the instructions here: http://www.fyneworks.com/jquery/star-rating/

I want to make it so I can submit the data to my database, which is very important.

Diego/others, assistance welcome please?
Do you want to see the entire page?

Cheers!
Jack

Jack Franklin · #2 (**permalink**) Feb 18th, 2008, 15:21

OK, my problem has changed. The question now is how to add it to the PHP while statement below?

PHP: Select all


while ($sites = mysql_fetch_array($query)) {
$siteid = $sites['site_id'];
$sitetitle = $sites['site_title'];
$siteurl = $sites['site_url'];
$siteimg = $sites['site_img'];
$username = $sites['user_name'];
$userurl = $sites['user_url'];
$accepteddate = $sites['accepted_date'];
$rating = $sites['rating'];
$times_rating = $sites['times_rated'];
$views = $sites['views'];
echo '<div class="site">';
echo '<h3>' . $sitetitle . '</h3>';
if ($siteimg == '') {
echo '<img src="noimgprovided.gif" alt="No Image Provided" height="140" width="140"  />';
} else {
echo '<img src=' . $siteimg . ' alt="' . $sitetitle . '" height="140" width="140" />';
}
echo '<p class="info">' . $accepteddate . ' || Designer: <a href="' . $userurl . '">' . $username . '</a> || Views: ' . $views . '</p>';
echo '<p class="rating">';
if ($rating == 0) {
echo 'Not Yet Rated';
} else {
$act_rating = $rating/$times_rating;
$final_rating = round($act_rating);
echo 'Rating:' . $final_rating . ' From ' . $times_rating . ' Ratings';
}
echo '</p>';
echo '</div>';
}

spinal007 · #3 (**permalink**) Mar 7th, 2008, 16:13

Hi Jack, I just came across this thread and I would have been able to help with the jQuery issue. I'm no help with PHP though...

Jack Franklin · #4 (**permalink**) Mar 7th, 2008, 18:17

Hey Diego,

Thanks, but I got it working

Jack

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Help With JQuery Star Rating.