Question

在过去的几个小时里，我一直在尝试通过Ajax将Json对象发送到我的Action，数据到达动作，但其值为null。

如果需要任何其他信息，我很乐意更新问题。

谢谢，伊兰

jQuery：

$(function () {
    $(".button-search").click(function () {
        var manufacturer= $("#Manufacturer").val();
        var model = $("#Model").val();
        var rentPrice = $("#RentPrice").val();
        var gear = $("#Gear").val();
        var year = $("#Year").val();

        $.ajax({
            type: "POST",
            url: "/Car/SearchResults",
            data: {
                Manufacturer: manufacturer,
                Model: model,
                RentPrice: rentPrice,
                Gear: gear,
                Year: year,
            },
            success: function (results) {
                alert("success");
            },
            error: function (error) {
                alert("error");
            }
        })
    });
})

动作：

    [HttpPost]
    public ActionResult SearchResults(CarSearchModel model)
    {
        var searchResults = CarHelper.CarSearch(model.Manufacturer, model.Model, model.RentPrice, model.Year, model.Gear);

        JsonResult result = new JsonResult();
        result.Data = searchResults;

        return result;
    }

类别：

public class CarSearchModel
    {
        [StringLength(4, ErrorMessage="Maximum 4 digits.")]
        public int Year { get; set; }
        public string Model { get; set; }
        public string Manufacturer { get; set; }
        public string Gear { get; set; }

        [DataType(DataType.Currency)]
        [Display(Name="Rent Price")]
        public int RentPrice { get; set; }

        [Display(Name="Delay Fee")]
        public int DelayFee { get; set; }

        [DataType(DataType.Date)]
        [Required]
        [Display(Name="Rent Date")]
        public DateTime RentDate { get; set; }

        [DataType(DataType.Date)]
        [Required]
        [Display(Name = "Return Date")]
        public DateTime ReturnDate { get; set; }

    }

Answer 1

你需要添加：

dataType: "json"


       $.ajax({
            type: "POST",
            url: "/Car/SearchResults",
            dataType: "json",
            data: {
                Manufacturer: manufacturer,
                Model: model,
                RentPrice: rentPrice,
                Gear: gear,
                Year: year,
            },
            success: function (results) {
                alert("success");
            },
            error: function (error) {
                alert("error");
            }
        })

Answer 2

尝试这个应该对你有用，你也需要打开你的开发工具（F12）并添加你得到的错误细节。

$(function () {
    $(".button-search").click(function () {
        var manufacturer= $("#Manufacturer").val();
        var model = $("#Model").val();
        var rentPrice = $("#RentPrice").val();
        var gear = $("#Gear").val();
        var year = $("#Year").val();

        var input = {
            Manufacturer: manufacturer,
            Model: model,
            RentPrice: rentPrice,
            Gear: gear,
            Year: year
        };

        $.ajax({
            type: "POST",
            dataType: 'json',
            contentType: "application/json; charset=utf-8",
            url: "/Car/SearchResults",
            data: JSON.stringify(input),
            success: function (results) {
                alert("success");
            },
            error: function (error) {
                alert("error");
            }
        })
    });
})

这可能会对您有所帮助：How to call web api method using jquery ajax in ASP.NET MVC 4

Answer 3

我尝试使用示例应用程序开发类似的代码，这里是类似Ajax结果的工作代码。注意：ActionResult应该是JsonResult

 [HttpPost]
 public JsonResult Index(Practice model)
 {
  // Whatever your logic
  return Json(new { msg = model.PracticeName }, JsonRequestBehavior.AllowGet);
  }

像这样的Ajax请求，

$(document).ready(function () {
        $('#btn').click(function (ev) {
            ev.preventDefault();
            $.ajax({
                url: '@Url.Action("Index","Practices")',
                type: 'Post',
                contenttype:'application/json; charset=utf-8',
                data: {
                    PracticeId: $('#PracticeName').val() ,
                    PracticeName:$('#PracticePercentage').val()
                },
                success: function (data) {
                    alert(data.msg);
                }
            });
        });
    });

我的示例视图就是这样，

@model MvcApplication3.Models.Practice

@using (Html.BeginForm()) {
    @Html.AntiForgeryToken()
    @Html.ValidationSummary(true)

    <fieldset>
        <legend>Practice</legend>

        <div class="editor-label">
            @Html.LabelFor(model => model.PracticeName)
        </div>
        <div class="editor-field">
            @Html.EditorFor(model => model.PracticeName)
            @Html.ValidationMessageFor(model => model.PracticeName)
        </div>

        <div class="editor-label">
            @Html.LabelFor(model => model.PracticePercentage)
        </div>
        <div class="editor-field">
            @Html.EditorFor(model => model.PracticePercentage)
            @Html.ValidationMessageFor(model => model.PracticePercentage)
        </div>

        <p>
            <input type="button" value="Create" id="btn" name="btn" />
        </p>
    </fieldset>
}

我的样本模型就像这里

 public class Practice
    {
        public int PracticeId { get; set; }
        public string PracticeName { get; set; }
        public string PracticePercentage { get; set; }
    }

希望这可以帮助您解决问题。

<强>更新

在您的方案中删除＆＃34;年＆＃34;的验证属性从ViewModel中再试一次它工作正常。还要从视图模型中重命名Model属性以进行冲突。

就是这样，

class CarViewModel  {  

public int Year { get; set; }

public string Manufacturer { get; set; }

public string Gear { get; set; }

public int RentPrice { get; set; }

[Display(Name = "Delay Fee")]
public int DelayFee { get; set; }

[DataType(DataType.Date)]
[Required]
[Display(Name = "Rent Date")]
public DateTime RentDate { get; set; }

[DataType(DataType.Date)]
[Required]
[Display(Name = "Return Date")]
public DateTime ReturnDate { get; set; 
}

Answer 4

我建议检查ModelState以查看是否存在任何类型转换错误。

我相信您会收到如下所示的错误。

The parameter conversion from type 'System.String' to type 'xxx' failed because
no type converter can convert between these types.

这可能是由于您的控制器参数名称与＆＃39;模型＆＃39;相同。财产名称。

Answer 5

我对你的代码做了一些改动，它对我有用。

<强>模型

public class CarSearchModel
    {
        public string Manufacturer {get;set;}
        public string Model {get;set;}
        public string RentPrice {get;set;}
        public string Year {get;set;}
        public string Gear {get;set;}
    }

行动方法

[HttpPost]
public ActionResult SearchResults(CarSearchModel model1)
{
    return Json(model1); 
}

Jquery Post

$(function(){
    $('.button-search').click(function(){

        var data1= {    
                        Manufacturer: 'Audi',
                        Model: 'Q8',
                        RentPrice: '100',
                        Gear: '4',
                        Year: '2001'

                };
            $.ajax({
                url: '@Url.RouteUrl(new{ action="SearchResults", controller="Home"})',
                data: JSON.stringify(data1),
                    type: 'POST',
                    dataType: 'json',
                    contentType: "application/json; charset=utf-8",
                    success: function(resp) {
                    alert(JSON.stringify(resp));
                    }
            });
        return false;
    });
});

您只需要对json参数进行字符串化。构建并运行，它将起作用。

POST Action始终通过Ajax接收null

5 个答案: