解决方案 »

  1.   


    <script type="text/javascript">           
    $(document).ready(function() { 
      window.sub=  $('#fdenglu').ajaxForm({ 
     beforeSubmit:  validate    // 提交前,验证
        }); 
    });function validate(formData, jqForm, options) {  
        var yonghuming1 = $('input[name=yonghuming]').fieldValue(); 
        var denglumima1 = $('input[name=denglumima]').fieldValue(); 
            
        if (!yonghuming1[0] || !denglumima1[0]) { 
            alert('用户名和密码不能为空'); 
            return false; 
        } 
           var queryString = $.param(formData); 
                    return true; 
    }
       </script> 
    HTML
    <form name="denglu" id="fdenglu" action="wz/loginYongHu" method="post"><div id="rightcontent"><div id="rightcontentn">
    <div id="yonghulable">
    <span>登录名:</span>
    </div>
    <div id="yonghuming">
    <input type="text" name="yonghuming" id="yonghuming" />
    </div>
    <div id="mimalable">
    <span id="denglum">登录密码:</span>
    <span id="wangjim">忘记登录密码</span>
    </div>
    <div id="mima">
    <input type="text" name="denglumima" id="denglumima" />
    </div>
    <div id="kongjian">
    </div>
    <div id="denglu">
    <a href="javascript:void(window.sub.submit())"><span>登&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;录</span></a>
    </div>
    <div id="zhuce">
    <span>免费注册</span>
    </div></div>
    </div>
    </form>  
      

  2.   

    http://blog.csdn.net/zzq58157383/article/details/7718956  看看这个怎么写的!