尝试使用 Ajax 调用控制器方法 MVC 获取数据我的代码已附

Trying to Get Data using Ajax call to Controller method MVC My code Attached

我在下拉值更改时调用 jquery 函数 jquery 方法是,

function MyFunction() {
    alert($('#DDlSurvey').val());
    $.ajax({
        url: "@Url.Action("GetSelectedQuestion", "ConductSurveyController")",
        data: { prefix: $('#DDlSurvey').val() },
    type: "GET",
    dataType: "json",
    success: function (data) {
        //  loadData(data);
        alert(data)
        alert("Success");
    },
    error: function () {
        alert("Failed! Please try again.");
    }
});
  //$('#YourLabelId').val('ReplaceWithThisValue');
}
</script>

我正在调用的函数,我收到了下拉值提醒

现在,我正在调用的函数是控制器“ConductSurveyController”中的“GetSelectedQuestion

方法就像,

[HttpPost]
public JsonResult GetSelectedQuestion(int prefix)
{
    List<SelectList> Questions=new List<SelectList>();

   //  Here "MyDatabaseEntities " is dbContext, which is created at time of model creation.
    SurveyAppEntities ObjectSur = new SurveyAppEntities();
       // Questions = ObjectSur.Surveys.Where(a => a.ID.Equals(prefix)).toToList();

我不认为这个方法正在调用,因为我收到错误

"Failed! Please try again"

来自我的脚本。

希望得到您的建议

谢谢

 var e = from q in ObjectSur.Questions
         join b in ObjectSur.SurveyQuestions on q.ID equals b.QuestionID where b.SurveyID.Equals(prefix)
         select q ;
         return new JsonResult { Data = e, JsonRequestBehavior = JsonRequestBehavior.AllowGet };
}

您控制器中的方法用 HttpPost 装饰,而在您的 ajax 中您指定的请求类型是 get 。你可以改变你的方法来得到这样的:

[HttpGet]
public JsonResult GetSelectedQuestion(int prefix)
{
}

或者在 Ajax 调用中将您的请求类型更改为 post:

$.ajax({
    url: "@Url.Action("GetSelectedQuestion", "ConductSurveyController")",
    data: { prefix: $('#DDlSurvey').val() },
    type: "Post",

此外 ControllerConductSurveyController 中是多余的,您需要将其删除并简单地称其为 ConductSurvey:

url: '@Url.Action("GetSelectedQuestion", "ConductSurvey")',

我认为您直接使用了控制器名称。你的 ajax 代码应该是这样的。

var PostData= { prefix: $('#DDlSurvey').val() }
var ajaxOptions = {
        type: "GET",
        url: '@Url.Action("GetSelectedQuestion", "ConductSurvey")',//Actionname, ControllerName
        data: PostData,
        dataType: "json",
        success: function (result) {
            console.log(result);
        },
        error: function (result) {

        }
};
$.ajax(ajaxOptions);