Return 列出 obj/variables 而不是模型中的全部

Question

如何配置我的代码，以便在我的控制器的 jsonresult 方法中只有 featureid 和 featurename 变量 return，而不是我模型中的所有对象 returning.

业务逻辑：

public static List<FeatureModel> LoadFeatures(){
string sql = @"SELECT FeatureId, FeatureName
                        FROM Feature";
return SqlDataAccess.LoadData<FeatureModel>(sql);}

型号：

public class FeatureModel{

public int FeatureId { get; set; }
public string FeatureName { get; set; }
public string FeatureDescription { get; set; }}

控制器：

public JsonResult ViewFeatures(){
var data = LoadFeatures();
List<FeatureModel> feat = new List<FeatureModel>();

foreach (var row in data)
{
    feat.Add(new FeatureModel
    {
        FeatureId = row.FeatureId,
        FeatureName = row.FeatureName,

    });
}
return Json(feat, JsonRequestBehavior.AllowGet);

}

JS/Ajax得到结果：

[{"FeatureId":0,"FeatureName":"Insights","FeatureDescription":null}]

Answer 1

一种简单的方法是使用 System.Linq

public JsonResult ViewFeatures(){
var data = LoadFeatures();
List<FeatureModel> feat = new List<FeatureModel>();

foreach (var row in data)
{
    feat.Add(new FeatureModel
    {
        ModuleId = row.ModuleId,
        FeatureName = row.FeatureName,

    });
}
return Json(feat.Select(x => new {x.FeatureId, x.FeatureName }), JsonRequestBehavior.AllowGet);

Return 列出 obj/variables 而不是模型中的全部

Return listed obj/variables rather than all in model

.net

c#

model-view-controller

visual-studio