在MVC 3.0中将JSON返回给Action
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我承认,我很困惑:
我正在尝试返回一个已转换为JSON的简单对象,如下所示:
viewModel.updateCoder = function (coder) {
var coderJson = ko.toJSON(coder);
var coderJsonString = ko.utils.stringifyJson(coderJson);
$.ajax({
url: \"provider/UpdateCoder\",
type: \'POST\',
dataType: \'text\',
data: coderJsonString,
contentType: \'text/csv\',
success: function () { alert(\"Updated!\"); }
});
我的RouteTable条目如下所示:
routes.MapRoute(
\"UpdateCoder\",
\"provider/UpdateCoder/{coderDTO}\", // URL with parameters
new { controller = \"Provider\", action = \"UpdateCoder\", coderDTO = UrlParameter.Optional }
);
我的Controler动作如下所示:
[AcceptVerbs(HttpVerbs.Post)]
public string UpdateCoder( string coderDTO )
{
var rslt = \"success\";
//var coder = coderDTO.CoderDTOToCoder();
return rslt;
}
我在UpdateCoder参数(字符串coderDTO)中得到的是一个null;
这是我的后备位置,我希望将JSON对象(coderJson)发送给操作,但是会收到错误消息:\“为此对象未定义无参数的构造函数。\”当我这样做时,参数类型如下:
[AcceptVerbs(HttpVerbs.Post)]
public string UpdateCoder( **CoderDTO coderDTO** )
{
var rslt = \"success\";
//var coder = coderDTO.CoderDTOToCoder();
return rslt;
}
以及:ValueProviderFactories.Factories.Add(new JsonValueProviderFactory());在Global.asax中
CoderDTO类如下所示:
public class CoderDTO
{
public Int32 Id { get; set; }
public String CoderCode { get; set; }
public String Sal { get; set; }
public String LName { get; set; }
public String FName { get; set; }
public String MI { get; set; }
public String Facility { get; set; }
public String Title { get; set; }
public Boolean? IsContract { get; set; }
public Boolean? IsInactive { get; set; }
public Boolean? IsDeleted { get; set; }
public String Comments { get; set; }
public String AlternateId { get; set; }
public int CasesCoded { get; set; }
public CoderDTO(Coder coder)
{
Id = coder.Id;
CoderCode = coder.CoderCode;
Sal = coder.Sal;
LName = coder.LName;
FName = coder.FName;
MI = coder.MI;
Facility = coder.Facility;
Title = coder.Title;
if (coder.IsContract != null) IsContract = coder.IsContract;
if (coder.IsInactive != null) IsInactive = coder.IsInactive;
if (coder.IsDeleted != null) IsDeleted = coder.IsDeleted;
Comments = coder.Comments;
AlternateId = coder.AlternateId;
}
public Coder CoderDTOToCoder()
{
var coder = new Coder
{
Id = Id,
CoderCode = CoderCode,
Sal = Sal,
LName = LName,
FName = FName,
MI = MI,
Facility = Facility,
Title = Title
};
coder.IsContract = IsContract ?? false;
coder.IsInactive = IsInactive ?? false;
coder.IsDeleted = IsDeleted ?? false;
coder.Comments = Comments;
coder.AlternateId = AlternateId;
return coder;
}
}
coderJsonString看起来像这样:
{\"Id\":201,\"CoderCode\":\"GP \",\"Sal\":null,\"LName\":null,\"FName\":null,\"MI\":null,\"IsContract\":false,\"IsInactive\":false,\"Comments\":null,\"CasesCoded\":0,\"isBeingEdited\":false}
真是漫长的一天!感谢您的帮助,我正在吃晚饭!!
没有找到相关结果
已邀请:
3 个回复
磨标烫徽啪
对于当前使用传递字符串的情况,我认为您应该这样称呼它:
因此,基本上,您将创建一个具有参数名称和值的对象并对其进行字符串化。在这种情况下,coderJson是双重编码的。
蹦吃舷弦
杭难插