dongye9228 2011-09-16 11:05
浏览 30
已采纳

Jquery,ajax和加载问题

This is the HTML part:

div id="messages">
   <div class="messages">


<?php if(isset ($unread)) { ?>
   <p>You have <?php echo $unread?> unread messages.</p>
    <?php } ?>

<?php if(isset ($messages)) { ?>
        <?php foreach ($messages as $msg){ ?>
      <div class="col_3">             
<?php     
      if($msg['read'] == 0){ echo 'Status of message: Unreaded';}
      elseif($msg['read'] == 1){echo 'Status of message: Readed';}
     echo "<p>Message from: $msg[name]</p>";
     echo "<p>Sender email: $msg[email]</p>";        
     echo "<p>Message: <br />$msg[message]</p>"; ?>
     <a href="#" class="delete" id="<?php echo $msg['id'] ?>">Delete message</a>
      </div>
 <?php } ?>                
<?php } ?><!-------- end of if $message-------->
       </div><!------ end of div .messages--------->
   </div><!------ end of div #messages--------->

and JQ:

  $(".delete").click(function() {    
    var commentContainer = $(this).parent();
    var id = $(this).attr("id");            
    var string = 'id='+ id ;
    $.ajax({   
        url: "<?php echo site_url('messages/delete') ?>",
        type: "POST",
        data: string,
        cache: false,
        success: function(){
            commentContainer.slideUp('600', function() {$(this).remove();
                $('.messages').fadeOut('2000', function(){$(this).remove();
                    $('#messages').load("<?php echo site_url('messages/show') ?>");
                }); 
            });               
        }   
    });
    return false;
});

Code is working, but when it comes to the load nothing is being shown. I did

load("<?php echo site_url('messages/show') ?>", function() {

alert('Load was performed.'); });

and there was an alert, and when I look page source I can see that the content has been changed, but it is not displayed.

  • 写回答

1条回答 默认 最新

  • douzhi1937 2011-09-16 11:12
    关注

    When you view the source of a page that has been loaded via AJAX, it will never update. You will need to inspect the DOM in order to see what has changed.

    The reason is because the content is not actually on the page and is being dynamically added to the page.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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