怎样用jquery的ajax异步提交表单数据

<form id="addForm" action="${pageContext.request.contextPath}/admin/saveAdd" method="post">    
  <input type="text" name="name" placeholder="请输入名字" />
  <input type="password" name="password" placeholder="密码"/>
 </form>

$("#submitAdd").click(function(){
     
  var targetUrl = $("#addForm").attr("action");    
  var data = $("#addForm").serialize();     
   $.ajax({ 
    type:'post',  
    url:targetUrl, 
    cache: false,
    data:data,  
    dataType:'json', 
    success:function(data){      
      alert('success');
    },
    error:function(){ 
     alert("请求失败")
    }
   })
     
})