我有一个 php web 应用程序项目......我想实现一个 javascript onclick 代码.. 例如：当用户点击跟随这篇文章...... MySQL 查询插入到用户遵循这个的数据库......所以页面被刷新，之后它将显示为 FOLLOWED .. 执行此操作所需的 javascript 代码是什么......或者是否有任何适合的示例？
这是一个示例代码
<div id="main_content">
<?php
if(isset($_GET['et']))
{
$et = $_GET['et'];
}
$result = mysql_query("select * from events where event_type =$et") ;
require_once('queries/category_extract.php') ;
?>
<div id="body_content">
<div id="event_tag">
<a><?php echo $cattitle["categ_title"]?>s</a>
</div>
<?php
while($row=mysql_fetch_array($result) )
{
require('queries/followers_extract.php')
?>
<div id="postpreview">
<div id="postpreview_up">
<div id="postpreview_up_left">
<div id="postpreview_up_left_left">
<a>Event Title :</a>
</div>
<div id="postpreview_up_left_right">
<a><?php echo $row["event_title"] ?></a>
</div>
</div>
<div id="postpreview_up_right">
<img src="images/read_more.gif" />
</div>
</div>
<div id="postpreview_bottom">
<div id="postpreview_bottom_left">
<div id="postpreview_bottom_left_left">
<a>Date :</a>
</div>
<div id="postpreview_bottom_left_right">
<a><?php echo $row["event_date"] ?></a>
</div>
</div>
<div id="postpreview_bottom_right">
<div id="postpreview_bottom_right_left">
<?php
if($follower['follower_id'] ==NULL){echo " <img src='images/follow_button.jpg' /> " ; }
else { echo " <img src='images/follow_closed.jpg' /> " ;}
?>
</div>
<div id="postpreview_bottom_right_right">
<?php
if($follower['follower_id'] !=NULL){ echo " <img src='images/following_button.jpg' /> " ; }
else { echo " <img src='images/following_Closed.jpg' /> " ;}
?>
</div>
<div id="postpreview_bottom_right_right">
<?php
if($follower['follower_id'] !=NULL){ echo " <img src='images/unfollow_button.jpg' /> " ; }
else { echo " <img src='images/unfollow_closed.jpg' /> " ;}
?>
</div>
</div>
</div>
</div>
<div id="splitter"></div>
<?php
}
?>
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