javascript执行php代码

Question

我有一个php web应用程序项目...我想实现一个javascript onclick代码..例如： 当用户点击后按照这个帖子... MySQL查询插入到数据库中用户跟着这个...所以页面刷新，之后它将显示为FOLLOWED .. 这是什么JavaScript代码需要做到这一点......或者是否有任何示例可能适合？

这里是一个示例代码

 <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 
    } 
    ?> 
    <!--End Of Post Preview--> 

    <!--End Of Post Preview--> 

Answer 1

OK Go的这个：http://www.9lessons.info/2009/04/exactly-twitter-like-follow-and-remove.html

你，因为我读学习的基础知识。