                        课程
                    
                        /前端开发
                        
                            /jQuery
                        
                        /jQuery基础(五)一Ajax应用与常用插件

为什么用$("#frmV").ajaxSubmit(options) ;显示不了数据。改成 $("#frmV").ajaxForm(options) ;才可以

    <form id="frmV" method="post" action="#">
            <div id="divtest">
                <div class="title">
                    <span class="fl">个人信息页</span> 
                    <span class="fr">
                        <input id="btnSubmit" type="submit" value="提交" />
                    </span>
                </div>
                <div class="content">
                    <span class="fl">用户名：</span><br />
                    <input id="user" name="user" type="text" /><br />
                    <span class="fl">昵称：</span><br />
                    <input id="nick" name="nick" type="text" />
                    <div class="tip"></div>
                </div>
            </div>
        </form>
        
        <script type="text/javascript">
         $(function(){
             var options = {
                 url:"http://www.imooc.com/data/form_f.php",
                 target:".tip"//返回的结果放在..元素之下
             }
             $("#frmV").ajaxForm(options) ; 
         })
         
        </script>