MVC里面如果直接将数据返回到前端页面，我们常用的方式就是用return view()；
那么我不想直接用razor语法，毕竟razor这玩意儿实在是太难记了，还不如写ajax对接来得舒服不是
那么我们可以这么做
1.定义ActionResult,返回json，标记属性可以采用HttpPost,也可以是用HttpGet，按自己的需求来使用
public ActionResult UpdateDownloadInJson(string deviceNames,string programNames) ???????{ ???????????string[] deviceName = deviceNames.Split(‘,‘); ???????????string[] programName = programNames.Split(‘,‘); ???????????List<DownloadViewModel> DownloadViewModelList = new List<DownloadViewModel>(); ???????????foreach (string tempDeviceName in deviceName) ???????????{ ???????????????var _deviceId=deviceInfoService.FindSingle<DeviceInfo>(r => r.DeviceName == tempDeviceName).Id; ???????????????foreach (string tempProgramName in programName) ???????????????{ ???????????????????int _programId = publishDetailService.Set<ProgramInfo>().Where(r => r.ProgramName == tempProgramName).FirstOrDefault().Id; ???????????????????var progress= publishDetailService.Set<DeviceMaterial>().Where(r => r.DeviceId == _deviceId && r.ProgramId == _programId).FirstOrDefault().DownProgress; ???????????????????DownloadViewModelList.Add(new DownloadViewModel ???????????????????{ ???????????????????????DeviceId= (int)_deviceId, ???????????????????????DeviceName = tempDeviceName, ???????????????????????ProgramName = tempProgramName, ???????????????????????DownloadProgress = (int)progress ???????????????????}); ???????????????} ???????????} ???????????return Json(new AjaxResult ???????????{ ???????????????Result = DoResult.Success, ???????????????RetValue = DownloadViewModelList ???????????}, JsonRequestBehavior.AllowGet); ???????}2.采用JsonResult，最主要拿来处理ajax请求
[HttpPost] ???????[HandlerAjaxOnly] ???????public JsonResult CheckLogin(string username, string password, string code) ???????{ ???????????UserManage.LoginResult result = this.HttpContext.UserLogin(username, password, code); ???????????if (result == UserManage.LoginResult.Success) ???????????{ ???????????????return Json(new AjaxResult { Result = DoResult.Success, DubugMessage = "登陆成功。" }); ???????????} ???????????else ???????????{ ???????????????return Json(new AjaxResult { Result = DoResult.Faild, DubugMessage = "登陆失败，" + result.ToString() }); ???????????} ???????}具体的区别后续补充，用法基本就是这样。
MVC中返回json数据的两种方式
原文地址：https://www.cnblogs.com/yinxuejunfeng/p/9685460.html