转-MVC中处理Json和JS中处理Json对象
2013-03-17 19:26 Echo. 阅读(138) 评论(0) 收藏 举报网上转来以便不时之需,Controller代码如下
public ActionResult TestString() { return Content("success"); } public ActionResult TestJson() { JsonResult json = new JsonResult { Data = new { Name = "zzl", Sex = "male", } }; return Json(json); } public JsonResult TestList() { List<User> userList = new List<User> { new User{Name="zzl",Email="bfyxzls@sina.com"}, new User{Name="zhz",Email="zhanghangzheng@sina.com"}, }; JsonResult json = new JsonResult { Data = userList }; return Json(json); //return Json(json, JsonRequestBehavior.AllowGet);//前台AJAX如果是GET用这句 //什么时候用GET请求呢,当我们直接在浏览器中输入网址时,其实就是一个GET请求 //如果我们直接输入/Home/TestList这个网址,它会提示我们下载这个JSON格式的文档 }
View代码如下
@{ Layout = null; } <!DOCTYPE html> <html> <head> <title>Index</title> <script src="http://www.cnblogs.com/Scripts/jquery-1.4.4.min.js" type="text/javascript"></script> <script language="javascript" type="text/javascript"> $(document).ready(function () { $("#btn1").click(function () { $.ajax({ url: "/Home/TestString", dataType: "text", cache: false, data: null, type: "POST", success: function (data) { $("#msg").html(data); } }); }); $("#btn2").click(function () { $.ajax({ url: "/Home/TestJson", dataType: "json", cache: false, data: null, type: "POST", success: function (data) { $("#msg").html(data.Data.Name); } }); }); $("#btn3").click(function () { $.ajax({ url: "/Home/TestList", dataType: "json", cache: false, data: null, type: "POST", success: function (data) { var msg = ""; for (var i = 0, length = data.Data.length; i < length; i++) { msg += "<DiV>Name:" + data.Data[i].Name + ",Email:" + data.Data[i].Email + "</div>"; } $("#msg").html(msg); } }); }); }); </script> </head> <body> <div> <input type="button" id="btn1" value="test return string" /> <br /> <input type="button" id="btn2" value="test return json" /> <br /> <input type="button" id="btn3" value="test return list" /> <div id="msg"> </div> </div> </body> </html>
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