将数据数组从Javascript传递到C#

将数据数组从Javascript传递到C#,c#,javascript,ajax,asp.net-mvc,C#,Javascript,Ajax,Asp.net Mvc,这是我的班级: public class ARecipe { public string picture { get; set; } public string title { get; set; } public int cookingTime { get; set; } public int preparationTime { get; set; } public string IngredientList { get; set; } publ

这是我的班级:

public class ARecipe
{
    public string picture { get; set; }
    public string title { get; set; }
    public int cookingTime { get; set; }
    public int preparationTime { get; set; }
    public string IngredientList { get; set; }
    public string ingredientsDescription { get; set; }
    public int nbPersons { get; set; }
    public string Category { get; set; }
    public string difficulty { get; set; }
    public double nbStars { get; set; }

}
我的Ajax调用:

var dico = { 
            picture: $("#fakeInput").val(),
            title : $("#title").val(),
            cookingTime : $("#cookingTime").val(),
            preparationTime : $("#preparationTime").val(),
            IngredientList : $("#ingredientListArea").val(),
            ingredientsDescription : $("#preparationArea").val(),
            nbPersons : parseInt($("#select-nb-Persons").val()),
            Category : $("#select-category").val(),
            difficulty: $("#select-difficulty").val(),
            nbStars : 4
        };

        $.ajax({
            url: "/AddRecipe/TempData",
            type: 'POST',
            success: function (e) {
                //success event
            },
            ///Form data
            data: JSON.stringify(dico),
            ///Options to tell JQuery not to process data or worry about content-type
            cache: false,
            contentType: false,
            processData: false
        });
以及接收数据的方法:

 [HttpPost]
  public ActionResult TempData(ARecipe recipe) {

     return Json("");
  }
我的Ajax调用会转到TempData方法,但是当我使用调试器分析参数“recipe”时,我注意到所有字段都是“null”

为什么?

你有解决办法吗


谢谢

您将以JSON的形式发送数据,但服务器希望以常规POST数据的形式发送数据。只需让
ajax
方法将其转换为常规POST请求,而不是将其强制转换为JSON:

///Form data
data: dico,

只要纠正这些问题:

 $.ajax({
        url: "/AddRecipe/TempData",
        type: 'POST',
        dataType: 'json',
        contentType: 'application/json',
        success: function (e) {
            //success event
        },
        ///Form data
        data: JSON.stringify(dico),
        ///Options to tell JQuery not to process data or worry about content-type
        cache: false,
    });

 [HttpPost]
 public JsonResult TempData(ARecipe recipe) {

 return Json("");
}

指定dataType:'json'和contentType:'application/json',您可以尝试一下
data:dico,