c# - 将动态 C# 对象序列化为 JSON

Question

我有一个 ASP.NET MVC 应用程序。此应用程序包括 C#、Razor 和 JSON.NET。我有一个如下所示的 C# 对象:

public class MyBlueprint
{
  public string Name { get; set; }

  public string Approach { get; set; }

  public object Value { get; set; }
}

此时，我有一个 MyBlueprint 对象列表，如下所示:

List<MyBlueprint> x = new List<MyBlueprint>();
{
  new MyBlueprint { Name="Bill", Approach="straight", Value=1 },
  new MyBlueprint { Name="Bill", Approach="straight", Value=7 },
  new MyBlueprint { Name="Bill", Approach="straight", Value=10 },
  new MyBlueprint { Name="John", Approach="straight", Value=120 },
  new MyBlueprint { Name="John", Approach="straight", Value=105 },
  new MyBlueprint { Name="John", Approach="straight", Value=150 },
  new MyBlueprint { Name="John", Approach="straight", Value=100 },
  new MyBlueprint { Name="Jill", Approach="curved", Value=1 },
  new MyBlueprint { Name="Jill", Approach="curved", Value=2 },
  new MyBlueprint { Name="Jill", Approach="curved", Value=5 },
  new MyBlueprint { Name="Jill", Approach="curved", Value=8 }
};

我需要将 MyBlueprint 对象集合展平为预定义的 JSON 结构。最终目标是让 JSON 在我看来如下所示:

var results = [
  { "name": "Bill", "approach":"straight", options: { begin: 1, end: 10 } },
  { "name": "John", "approach":"straight", options: { begin: 100, end: 150 } },
  { "name": "Jill", "approach":"curved", options: { chosen: [1, 2, 5, 8] } }
];

options 参数根据 approach 的值具有不同的属性。上面的 JSON 显示了两种不同的结构:curved 和 straight。我还会有更多。因此，我试图找出一种动态方法，而无需创建大量模型类。目前，我有以下内容:

var names = new[] { "Bill", "John", "Jill" };
foreach (var name in names)
{
  var blueprints = x.Where(y => y.Name == name).Select(z => z.Value.ToString());    
}

我知道如何获取曲线集的选择值列表。我知道如何获取 straight 集的 begin 和 end 值。然而，一旦我拥有它们，我不知道如何处理它们。我试图弄清楚如何创建一个动态 C# 对象，我可以将其序列化为 JSON 并将其放入 View 中。我缺少什么？我感觉有点被困住了。

Answer 1

我建议你必须做以下事情。

创建类。

public class Options
{
    public Options()
    {
    }

    public Options(List<int> d,string approach)
    {
       if (approach == "straight")
       {
          begin = d.Min();
          end = d.Max();
       }
       else if (approach == "curved")
       {
           choosen = d.Select(cc => cc).ToArray();
       }
    }

    public int? begin { get; set; }
    public int? end { get; set; }
    public int[] choosen { get; set; }
}

和你的 linq 查询

 var result = (from item in x
                          group item by new { item.Name, item.Approach } into g
                          select new
                          {
                              name = g.Key.Name,
                              approach = g.Key.Approach,
                              options = new Options(g.Select(cc => Convert.ToInt32(cc.Value)).ToList(), g.Key.Approach)
                          }).ToList();

            string value = Newtonsoft.Json.JsonConvert.SerializeObject(result, new Newtonsoft.Json.JsonSerializerSettings() { NullValueHandling = Newtonsoft.Json.NullValueHandling.Ignore });