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。
原文地址:https://www.cnblogs.com/yinxuejunfeng/p/9685460.html
时间: 2024-11-05 11:56:20