Question

我的ajax代码就在这里，

成功功能总是显示两次。

如果我在数据库中有两条记录意味着它会两次检索这两行，三行意味着检索两次

<script type="text/javascript">
            $(document).ready(function()
            {
                $('form#form1').submit(function(e) {
                    var form = $(this);
                    var srpid = form.find('#srpid').val();
                    //$('#cmd_'+srpid).html(" ");
                    e.preventDefault();
                    $.ajax({
                        'type': "POST",
                         url: "<?php echo base_url('pages/post/comments'); ?>",
                        'data': form.serialize(), // <--- THIS IS THE CHANGE
                        //dataType: "html",
                        'success': function(data){
                            $('#cmd_'+srpid).html(data);
                        },
                        //error: function() { alert("Error posting feed."); }
                   });
                  return false;
                });     
            });
            </script>


        <div class="post-scroll" id="cmd_<?php echo $post_id; ?>">
    <?php 
    $command = $this->db->order_by('comment_date','DSC')->where('post_id',$post_id)->get('sr_post_comment')->result();
    foreach($command as $cmd){ 
                    $usercmdview = $this->db->get_where('users',array('id' => $cmd->user_id))->result();
    ?>
     <div class="post-cnt"  style="float:left;">
        <div class="post-img">
             <img src="<?php echo $usercmdview[0]->image; ?>" />
        </div>
        <div class="post-name">
            <p><?php echo $usercmdview[0]->firstname." "; ?><br /><span><?php echo  $cmd->comment_date; ?></span></p> 
        </div>
        <div class="post-cmd">
             <p class="cmdlist_<?php echo $cmd->comment_id; ?>"><?php echo $cmd->comments; ?></p>
        </div>
     </div>
 <?php } ?>
</div>

Answer 1

您在＆＃34;提交＆＃34;定义了ajax请求行动，但你没有停止表格提交。这意味着当您点击＆＃34;提交＆＃34;按钮，将执行以下操作。

ajax将数据提交给服务器
表单也提交，它将刷新页面。

＆＃13;

<script type="text/javascript">
    $(document).ready(function(){
        $('form#form1').submit(function(e) {
            ...
            return false;
        });     
    });
</script>

＆＃13;

Ajax成功函数在codeigniter中重复加载div

1 个答案: