1.使用ajax将前台数据传递到后台(将字符串messName传递到后台,接收返回的json数据,并进行处理)
function getWarmMessages() {$.ajax({type: "post",url: "Warning/Detailed",data: {messName: messName},success: function (result) {var jsonResult = JSON.parse(result);alert(jsonResult[0].id);}});
}
2.WarningController.cs中新建Detailed方法。(访问数据库,将获取到的数据转换为json格式返回)
public JsonResult Detailed(string messName){var bwr = from b in db.BSB_Warm_Recordswhere b.StationName == messNameselect b;var bwrList = bwr.ToList();DataContractJsonSerializer json = new DataContractJsonSerializer(bwrList.GetType());string szJson = "";using (MemoryStream stream = new MemoryStream()){json.WriteObject(stream, bwrList);szJson = Encoding.UTF8.GetString(stream.ToArray());}return Json(szJson);}