如何通过 Jquery (Ajax) 将复杂类型发送到 MVC2 控制器答案

【问题标题】：How to send complex types to MVC2 controller via Jquery (Ajax )如何通过 Jquery (Ajax) 将复杂类型发送到 MVC2 控制器
【发布时间】：2011-02-22 19:42:30
【问题描述】：

我正在向我的控制器发布一个 Jquery Json 序列化对象，但并非所有数据都被传递。其中一个成员是一个复杂类型，也是 Json 序列化的。这是没有通过控制器的那个。

这是我通过 Ajax 帖子传递给我的控制器的类。请注意复杂类型 RoutingRuleModel。

SourceCodeModel.cs：

[Serializable]
public class SourceCodeModel
{
    public string SourceCode { get; set; }
    public bool IsActive { get; set; }
    public string LastChangedBy { get; set; }
    public string LocationCode { get; set; }
    public string Vendor { get; set; }
    public RoutingRuleModel RuleModel { get; set; }
}

RoutingRuleModel.cs：

[Serializable]
public class RoutingRuleModel
{
    public string AdaStuInfoSysId { get; set; }
    public string AdaInitials{ get; set; }
    public string LocationCode { get; set; }
    public string Vendor { get; set; }
    public string RuleName { get; set; }
    public string RuleStatus { get; set; }
    public int RuleId { get; set; }
}

这是我在 JavaScript 中构建模型的方式：

getSourceCodeModel = function (sourceCode, isActive, lastChangedBy, locationCode,   ruleModel) {
    // Retrieves a SourceCodeModel object that can be JSON-serialized
    return ({
                    SourceCode: sourceCode,
        IsActive: isActive,
        LastChangedBy: lastChangedBy,
        LocationCode: locationCode,
        Vendor: ruleModel.Vendor,

        RuleModel: [{ AdaStuInfoSysId: ruleModel.AdaStuInfoSysId, AdaInitials: ruleModel.AdaInitials, LocationCode: ruleModel.locationCode, 
            Vendor: ruleModel.Vendor, RuleName: ruleModel.RuleName, RuleStatus: ruleModel.RuleStatus, RuleId: ruleModel.RuleId}]
    });
};

这是我的 JQuery Ajax 调用：

$.ajax({
            type: "POST",
            url: "/LeadRoutingConsole/VendorLeadRouting/PostSourceCode",
            data: sourceCodeModel,
            datatype: "json",
            success: Commit_success,
            error: Commit_error,
            complete: function (jqXHR) { }
        });

这是我的控制器的动作方法：

[HttpPost]
    public JsonResult PostSourceCode(SourceCodeModel model)
    {
        // perform the save op
        var viewModel = new SourceCodesViewModel();
    viewModel.PostSourceCode(model);
        return Json(model);
    }

问题：SourceCodeModel 包含正确的值，除了它的复杂成员：RuleModel，它作为具有默认（null 或 0）值的 RoutingRuleModel 返回。

【问题讨论】：

标签： jquery asp.net ajax json asp.net-mvc-2

【解决方案1】：

您正在尝试以RuleModel 发送集合，而这不是集合属性。所以尝试这样（删除RuleModel 属性定义周围的方括号[]）：

getSourceCodeModel = function (sourceCode, isActive, lastChangedBy, locationCode,   ruleModel) {
    return ({
        SourceCode: sourceCode,
        IsActive: isActive,
        LastChangedBy: lastChangedBy,
        LocationCode: locationCode,
        Vendor: ruleModel.Vendor,
        RuleModel: { 
            AdaStuInfoSysId: ruleModel.AdaStuInfoSysId, 
            AdaInitials: ruleModel.AdaInitials, 
            LocationCode: ruleModel.locationCode, 
            Vendor: ruleModel.Vendor, 
            RuleName: ruleModel.RuleName, 
            RuleStatus: ruleModel.RuleStatus, 
            RuleId: ruleModel.RuleId
        }
    });
};

也永远不要像在 javascript 中那样对 url 进行硬编码。在处理 url 时始终使用 Url 助手，否则在部署应用程序时可能会遇到意外：

$.ajax({
    type: 'POST',
    url: '<%= Url.Action("PostSourceCode", "VendorLeadRouting") %>',
    data: sourceCodeModel,
    dataType: 'json',
    success: Commit_success,
    error: Commit_error,
    complete: function (jqXHR) { }
});

还要注意，该参数被称为 dataType: 'json' 而不是 datatype: 'json'，这对于 javascript 等区分大小写的语言很重要。

【讨论】：

感谢您的回复和关于不要硬编码网址的建议。我做了你建议的修复，它仍然不起作用！这来自 FireBird：'SourceCode=333www&IsActive=1&LastChangedBy=&LocationCode=AICA&Vendor=STUSCOUT&RuleModel%5BAdaStuInfoSysId%5D=&RuleModel%5BAdaInitials%5D=&RuleModel%5BLocationCode%5D=AICA&RuleModel%5BVendor%5D=STUSCOUT&RuleModel%5BRuleName%5D=&RuleModel% %5D=&RuleModel%5BRuleId%5D'

【解决方案2】：

您可以在控制器操作上编写 CustomFilterAttribute 并反序列化 CustomFilterAttribute 类中的 json 对象。比如……

public class JsonFilter : ActionFilterAttribute
{
    public string Param { get; set; }        
    public Type JsonType { get; set; }
    public override void OnActionExecuting(ActionExecutingContext filterContext)
    {
       DataContractJsonSerializer serializer = new DataContractJsonSerializer(typeof(SourceCodeModel));
       filterContext.HttpContext.Request.InputStream.Position = 0;
       SourceCodeModel result = serializer.ReadObject(filterContext.HttpContext.Request.InputStream)
              as SourceCodeModel;
       filterContext.ActionParameters[Param] = result;
    }
}

和控制器动作...

[HttpPost]
[JsonFilter(Param = "sourceCodeModel", JsonType = typeof(SourceCodeMode))]
public JsonResult PostSourceCode(SourceCodeModel model)
{...}

【讨论】：