CSS实现星级评分III

2009-12-16 23:34| 发布者: admin| 查看: 48| 评论: 0|原作者: 韩菱纱

首先是实现的原理
从上一个css实现星级评分I 、II，可是看出，只要能识别onclick和将数据记录至数据库中存储，然后从数据库中调用出数据进行计算就
可以得到当前的评分均值——当前的分值。

1.下面是建立数据库的sql

CREATE TABLE ratings(
id INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
total_votes INT NOT NULL,
情缘.教程_网 [http://Www.XiuGoo.Com]
total_value INT NOT NULL,
which_id INT NOT NULL,
which_table VARCHAR(255),
used_ips LONGTEXT NULL
);
2.参数文件引用

require("connectDB.php");
require("closeDB.php");
require("tableName.php");
require("openDB.php");
?>

3.显示投票程序和更新投票数据程序

$rating_posted=$_GET['vote'];//pased variable by the the stars value
$id=$_GET['id'];
$query=mysql_query("SELECT total_votes, total_value, used_ips FROM $tableName WHERE id='".$id."' ")or die(" Error: ".mysql_error());
$numbers=mysql_fetch_assoc($query);
$checkIP=unserialize($numbers['used_ips']);
$count=$numbers['total_votes'];//how many votes total
$current_rating=$numbers['total_value'];//total number of rating added together and stored
$sum=$rating_posted+$current_rating;// add together the current vote value and the total vote value
$tense=($count==1) ? "vote" : "votes";//plural form votes/vote
$voted=@mysql_fetch_assoc(@mysql_query("SELECT title FROM $tableName WHERE used_ips LIKE '%".$_SERVER['REMOTE_ADDR']."%' AND id='$id' ")); //Pattern match ip:suggested by Bramus! http://www.bram.us/ - this variable searches through the previous ip address that have voted and returns true or false

if($voted){
echo "

".
"

Current rating.

".
"

Rating: ".@number_format($current_rating/$count,2)." {".$count." ".$tense." cast}
You have previously voted.

";//show the current value of the vote with the current numbers
}else{

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

if($sum==0){
$added=0;//checking to see if the first vote has been tallied
}else{
$added=$count+1;//increment the current number of votes
}

if(is_array($checkIP)){
array_push($checkIP,$_SERVER['REMOTE_ADDR']);//if it is an array i.e. already has entries the push in another value
}else{
$checkIP=array($_SERVER['REMOTE_ADDR']);//for the first entry
}

$insert=serialize($checkIP);
mysql_query("UPDATE $tableName SET total_votes='".$added."', total_value='".$sum."', used_ips='".$insert."' WHERE id='".$_GET['id']."'");

echo "

Rating: ".@number_format($sum/$added,2)." {".$added." ".$tense." cast} Thank you for your vote!

";//show the updated value of the vote
}else{
?>
4.访问者评分程序

CSS: Star Rater Example

How clear was this tutorial?

Current rating

" title="Rate this 1 star out of 5" class="one-star">1

" title="Rate this 2 stars out of 5" class="two-stars" >2

" title="Rate this 3 stars out of 5" class="three-stars" >3

" title="Rate this 4 stars out of 5" class="four-stars" >4

" title="Rate this 5 stars out of 5" class="five-stars" >5

5.最新评分结果提示

echo "

Rating: ".@number_format($sum/$count,2)." {".$count." ".$tense." cast}

";//show the current updated value of the vote
} // end isset get vote
} //end voted true, false
?>
下一步是将结果记入数据库，现在没有时间去研究了，请大家等待下一篇文章或者去远出处阅读！
原文：Creating a star rating system part (3)
链接：http://slim.climaxdesigns.com/tutorial.php?id=3
版权：版权归原作者所有，翻译文档版权归本人|greengnn，转载请注明出处www.jluvip.com/blog
我们用php来实现
先看看效果：http://gnn.80x86.cn/starrating

收藏分享邀请

		自动登录	找回密码
密码			注册

CSS实现星级评分III

最新评论

相关分类