在ajax jquery中调用ajax

时间:2013-11-30 06:32:32

标签: php jquery ajax

我需要有关如何在另一个ajax中恢复和ajax结果的帮助

这是我的HTML

         <div id="res_videoclick">
           <iframe class="laptopvideo"  src="http://www.ustream.tv/embed/recorded/37579887?v=3&amp;wmode=direct" scrolling="no" frameborder="0" style="border: 0px none transparent;">
          </iframe>
          </div>


         <ul id="videomenu">
            <li><a class="videocategory selected"  href="HEARTLINK">HEARTLINK</a></li>
            <li><a class="videocategory" href="WEIGHT LOSS: LEVEL 1">WEIGHT LOSS: LEVEL 1</a></li>
            <li><a class="videocategory" href="GROCERY STORE FINDS">GROCERY STORE FINDS</a></li>
            <li><a class="videocategory" href="TESTIMONIES">TESTIMONIES</a></li>
         </ul>

        <ul id="mycarousel"  class="jcarousel-skin-tango"> 
         <?php
        $videos_query=mysql_query("select * from videos where category='HEARTLINK' order by category, title  ASC" );
while($vd=mysql_fetch_array($videos_query))
                {
                    $videosid   = $vd['videosid'];
                    $title  = $vd['title'];
                    $link   = $vd['link'];
                    $linktwo    = substr($link, 31, 2);
                    $linkfive   = substr($link, 31, 5);
                    $linkfull   = substr($link, 31, 8);
                    $category   = $vd['category']; 
                    $description    = $vd['description'];   
                ?>
                <li><a class="videoclick" id="<?php echo $linkfull ;?>"><img style=" width:180px;;" src="http://static-cdn1.ustream.tv/videopic/0/1/<?php echo $linktwo ;?>/<?php echo $linkfive ;?>/<?php echo $linkfull ;?>/1_15835591_<?php echo $linkfull ;?>_320x240_b_1:2.jpg" alt="<?php echo $title;?>" title="<?php echo $title;?>" /></a></li>       

                <?php
                    }
                    ?>
        </ul>

AND HERE 2 AJAX REQUEST

 $(".videocategory").click(function() 
    {  

 $('#mycarousel').html('<small><img src="/travismartin/files/images/loading.gif" width="20" align="absmiddle"><br clear="all">Checking Database...</small>');
         var videocategory = $(this).attr('href');
     $.ajax({
            type: "POST",  
            url: "/travismartin/files/ajax/ajax_change.php",  
            data: {videocategory:videocategory},
            success: function(result)
            {  
                $('#mycarousel').html(result);
            }
        });   
});   

 $(".videoclick").bind('click',function(event) 
    {  

 $('#res_videoclick').html('<small><img src="/travismartin/files/images/loading.gif" width="20" align="absmiddle"><br clear="all">Checking Database...</small>');
         var videoclick = $(this).attr('id');
     $.ajax({
            type: "POST",  
            url: "/travismartin/files/ajax/ajax_change.php",  
            data: {videoclick:videoclick},
            success: function(result1)
            {  


    $('#res_videoclick').html(result1);
                    $('html, body').stop().animate({
                        scrollTop: $("#res_videoclick").offset().top
                    }, 500);
                      event.preventDefault();
                }
            });  

当一个类视频摄像机点击一个li a class = videoclick将出现..并且这完美...但是当我点击videoclick上的结果时... res_videoclick结果不会出现这个错误吗?请提前帮助我谢谢

2 个答案:

答案 0 :(得分:0)

使用此

$(document).on('click',".videoclick",function(event) 

而不是

$(".videoclick").bind('click',function(event) 

答案 1 :(得分:0)

对逻辑进行一些修改:

$('#mycarousel').hide();

$(".videocategory").click(function() 
{  

   var videocategory = $(this).attr('href');  

   $.ajax({
        type: "POST",  

        url: "/travismartin/files/ajax/ajax_change.php",  

        beforeSend: function() {
           $('#mycarousel').html('<small><img src="/travismartin/files/images/loading.gif" width="20" align="absmiddle"><br clear="all">Checking Database...</small>');
        },

        complete: function() {
           $('#mycarousel').hide();
        },

        data: {videocategory:videocategory},

        success: function(result)
        {  
            $('#mycarousel').html(result);
        }

    }); 

});