Question

不从C＃Controller Action返回AJAX成功我尝试了很多可用的选项，但它仍然没有将数据返回给AJAX成功可能是什么原因？我想返回列表Listemr_t_Immunization从控制器操作到我的成功AJAX调用数据我应该怎么做

请参阅以下代码谢谢

AJAX请求

$.ajax(
             {
                 url: '@Url.Action("getmedi", "Immunization")',
                 type: 'Post',
                 async: false,
                 contentType: 'application/json',
                 dataType: "json",
                 data: JSON.stringify({ "patient2": patient2}),
                 success: function (data) {
                     debugger
                     $(data).each(function(index, element){
                         $('#first > tbody').append('<tr><td></td><td></td><td></td><td></td><td></td><td></td></tr>');
                         $('#first > tbody > tr:last-child > td:nth-child(1)').append('<input id='+element.ImmunizationId+' value='+element.VaccineName+'>');
                         $('#first > tbody > tr:last-child > td:nth-child(2)').append('<input id='+element.Immunization_Id+' value='+element.DateGiven+'>');
                         $('#first > tbody > tr:last-child > td:nth-child(3)').append('<input id='+element.Immunization_Id+' value='+element.Route+'>');



                     })
                     $("input[name='VaccineName']").attr('disabled', 'disabled');
                     $("input[name='DateGiven']").attr('disabled', 'disabled');
                     $("input[name='Route']").attr('disabled', 'disabled');

                 },
                 error: function (textStatus, errorThrown)
                 {
                     debugger
                 }

             });

控制器操作

[HttpPost]
public JsonResult getmedi(int patient2)
        {
            if (patient2.ToString() != "")
            {
                string roster = objOrgCont.getmedi(patient2);

                JavaScriptSerializer ser = new JavaScriptSerializer();

                emr_t_Immunization erx = (emr_t_Immunization)ser.Deserialize(roster, typeof(emr_t_Immunization));

                List<emr_t_Immunization> Listemr_t_Immunization = db.emr_t_Immunization.Where(Allemr_t_Immunization => Allemr_t_Immunization.Patient_Id == patient2).ToList();

                ///List<emr_t_Medical_History> Listemr_t_Medical_History2 = (from Allemr_t_Medical_History in db.emr_t_Medical_History where Allemr_t_Medical_History.Mr_No == Mr_No select Allemr_t_Medical_History).ToList();

                if (erx != null)
                {
                    //return Json(new { success = true, for_Date = erx.Med_Date, for_Name = erx.Name, for_Active = erx.Active, for_Resolved = erx.Resolved, for_Comments=erx.Comments });
                    return Json(new { Listemr_t_Immunization, JsonRequestBehavior.DenyGet });
                }

            }

            return Json(new { success = false });

        }

我还试图在返回之前对其进行字符串化并显示以下错误

我已经习惯了 return JsonConvert.SerializeObject(Listemr_t_Immunization);

现在它给我错误

使用类型检测到自引用循环＆＃39; System.Data.Entity.DynamicProxies.as_t_appointment_305685F58BEE75BAC95975F9457412A0DE58BB59D3EDBD1155C7DB5E21A7BA66＆＃39 ;. 路径＆＃39; [0] .erx_t_patient.as_t_appointment [0] .erx_t_city.as_t_appointment＆＃39;

Answer 1

只需快速浏览一下，看起来您可以将失败返回到其他语句中以保护它们不会被解雇，除非您的某个条件失败，而不是仅仅自行浮动。

您的回叫每次都失败的问题是什么？如果是这样可能是原因。

[HttpPost]
public JsonResult getmedi(int patient2)
    {
        if (patient2.ToString() != "")
        {
            string roster = objOrgCont.getmedi(patient2);

            JavaScriptSerializer ser = new JavaScriptSerializer();

            emr_t_Immunization erx = (emr_t_Immunization)ser.Deserialize(roster, typeof(emr_t_Immunization));

            List<emr_t_Immunization> Listemr_t_Immunization = db.emr_t_Immunization.Where(Allemr_t_Immunization => Allemr_t_Immunization.Patient_Id == patient2).ToList();

            ///List<emr_t_Medical_History> Listemr_t_Medical_History2 = (from Allemr_t_Medical_History in db.emr_t_Medical_History where Allemr_t_Medical_History.Mr_No == Mr_No select Allemr_t_Medical_History).ToList();

            if (erx != null)
            {
                //return Json(new { success = true, for_Date = erx.Med_Date, for_Name = erx.Name, for_Active = erx.Active, for_Resolved = erx.Resolved, for_Comments=erx.Comments });
                return Json(new { Listemr_t_Immunization, JsonRequestBehavior.DenyGet });
            } else 
            {
               return Json(new { success = false });
            }
        } else

        {
          return Json(new { success = false });
        }

    }

另外在旁注中，处理API-ish成功和失败的更好方法是使用HTTP错误代码。

 Response.StatusCode = (int)HttpStatusCode.Created;
 return Json(new { results });

修改

在您的请求中发回一个列表。

这一行需要改变：

List<emr_t_Immunization> Listemr_t_Immunization = db.emr_t_Immunization.Where(Allemr_t_Immunization => Allemr_t_Immunization.Patient_Id == patient2).ToList();

我猜这个db是你的上下文对象，如果是这样，用以下代码替换该行：

var results = db.emr_t_Immunization.Where(Allemr_t_Immunization => Allemr_t_Immunization.Patient_Id == patient2).ToList()

我不是100％确定erx是什么，但是如果你检查你有结果，那么将其更改为。

  if (results.Any())
    {
     Response.StatusCode = (int) HttpStatusCode.Created;
     return Json(new { results });
    } else 
    {
      Response.StatusCode = (int)HttpStatusCode.BadRequest;
      return Json("Failed");
    }

所以你的代码都会读到：

 [HttpPost]
     public JsonResult getmedi(int patient2)
      {
          if (patient2.ToString() != "")
          {
            string roster = objOrgCont.getmedi(patient2);
            JavaScriptSerializer ser = new JavaScriptSerializer();
            emr_t_Immunization erx =(emr_t_Immunization)ser.Deserialize(roster, typeof(emr_t_Immunization));
            var results = db.emr_t_Immunization.Where(Allemr_t_Immunization     => Allemr_t_Immunization.Patient_Id == patient2).ToList()
            /// You can make that query easier to read by having
            /// var results = db.emr_t_Immunization.Where(a => a.Patient_Id == patient2).ToList()
        if (results.Any())
         {
          Response.StatusCode = (int) HttpStatusCode.Created;
          return Json(new { results });
         } else 
         {
           Response.StatusCode = (int)HttpStatusCode.BadRequest;
           return Json("Failed");
         }
      } else
      {
        Response.StatusCode = (int)HttpStatusCode.BadRequest;
        return Json("Failed");
      }

    }

不从C＃Controller Action返回AJAX成功

1 个答案: