一:Jquery如何使用JSON传递多个参数给Controller中Action方法

案例一:ajax传递一个基本参数给action

<script type="text/javascript">
    $(document).ready(function () {
        var postData = { userId: 4 };
        var url = "@Url.Action("GetUserName")";
         $.ajax({
             async: false,
             type: "POST",
             url: url,
             data: postData,
             cache: false,
             global: false,
             dataType: ‘json‘,
             success: function (data) {
               
             }
         });
    });
</script>

后台Controller:
        public ActionResult GetUserName(int userId)
        {
            JsonResult jt = Json(new { success = true });
            jt.ContentType = "text/html";
            return jt;
        }

二:Jquery如何处理Controller返回的List集合对象

三:相关总结

MVC4与JSON交互的知识总结,布布扣,bubuko.com

MVC4与JSON交互的知识总结