繁体   English   中英

JavaScript执行PHP代码

[英]javascript executing php code

我有一个php Web应用程序项目...,我想实现一个javascript onclick代码..例如:当用户单击时,请遵循此信息... MySQL查询将插入到用户遵循此信息的数据库中。因此,该页面刷新后,它将显示为FOLLOWED ..进行此操作所需的javascript代码是什么?或者有没有合适的示例?

这是示例代码

     <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-->

好的,可以这样做: http : //www.9lessons.info/2009/04/exactly-twitter-like-follow-and-remove.html

您必须阅读我所读的基础知识。

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM