使用 Newtonsoft.Json.Schema 有没有办法拦截“定义”名称的生成？

Question

鉴于我有以下类型：

public class MyClass
{
    public MyGenericClass<bool> Gen { get; set; }
}

public class MyGenericClass<T>
{
    [Required]
    public T Prop { get; set; }
}

我像这样设置模式生成器

var generator = new JSchemaGenerator
{
     ContractResolver = new CamelCasePropertyNamesContractResolver(),
     SchemaIdGenerationHandling = SchemaIdGenerationHandling.None
};

generator.GenerationProviders.Add(
     new StringEnumGenerationProvider { CamelCaseText = true });

我得到以下架构

{
  "definitions": {
    "MyGenericClass<Boolean>": {
      "type": [
        "object",
        "null"
      ],
      "properties": {
        "prop": {
          "type": "boolean"
        }
      },
      "required": [
        "prop"
      ]
    }
  },
  "type": "object",
  "properties": {
    "gen": {
      "$ref": "#/definitions/MyGenericClass<Boolean>"
    }
  },
  "required": [
    "gen"
  ]
}

这很好，只是我需要与之共享架构的服务不喜欢尖括号“”。

有没有办法拦截定义名称的创建，以便我可以去掉 ？请注意，其他 SchemaIdGenerationHandling 也不起作用。

有点像

public class GenericSchemaProvider : JSchemaGenerationProvider
{
    public override bool CanGenerateSchema(JSchemaTypeGenerationContext context)
    {

        return context.ObjectType.IsGenericType && !context.ObjectType.IsAssignableFrom(typeof(IEnumerable));
    }

    public override JSchema GetSchema(JSchemaTypeGenerationContext context)
    {
        return CanGenerateSchema(context) ? CreateSchemaWithCustomId(context, context.Required) : null;
    }

    private static JSchema CreateSchemaWithCustomId(JSchemaTypeGenerationContext context, Required required)
    {
        var schema = context.Generator.Generate(context.ObjectType, required != Required.Always);
        schema.Id = new Uri(TypeName(context.ObjectType), UriKind.Relative);
        return schema;
    }

    private static string TypeName(Type type, string acc = "")
    {
        var name = type.Name;
        var index = name.IndexOf('`');
        return index == -1 ? acc+name : TypeName(type.GetGenericArguments()[0],acc+name.Substring(0, index));
    }
}

几乎可以工作

{
  "definitions": {
    "MyGenericClassNullableDouble": {
      "$id": "MyGenericClassNullableDouble",
      "type": [
        "object",
        "null"
      ],
      "properties": {
        "prop": {
          "type": "number"
        }
      },
      "required": [
        "prop"
      ]
    },
    "MyGenericClassDouble": {
      "$id": "MyGenericClassDouble",
      "type": [
        "object",
        "null"
      ],
      "properties": {
        "prop": {
          "type": "number"
        }
      },
      "required": [
        "prop"
      ]
    },
    "MyGenericClassBoolean": {
      "$id": "MyGenericClassBoolean",
      "type": [
        "object",
        "null"
      ],
      "properties": {
        "prop": {
          "type": "boolean"
        }
      },
      "required": [
        "prop"
      ]
    }
  },
  "type": "object",
  "properties": {
    "genBool": {
      "$ref": "MyGenericClassBoolean"
    },
    "genDouble": {
      "$ref": "MyGenericClassDouble"
    },
    "genNulableDouble": {
      "$ref": "MyGenericClassNullableDouble"
    }
  }
}

但是这些引用并不是对定义的正确引用。

Answer 1

我相信你在正确的轨道上，我认为你只需要提供自己的JSchemaGenerationProvider 实现，而不是使用StringEnumGenerationProvider。类似creating a custom provider:

public class FormatSchemaProvider : JSchemaGenerationProvider
{
    public override JSchema GetSchema(JSchemaTypeGenerationContext context)
    {
        // customize the generated schema for these types to have a format
        if (context.ObjectType == typeof(MyGenericClass<bool>))
        {
            return CreateSchemaWithFormat(
               context.ObjectType, context.Required, "MyGenericClass");
        }
        // use default schema generation for all other types
        return null;
    }

    private JSchema CreateSchemaWithFormat(Type type, Required required, string format)
    {
        JSchemaGenerator generator = new JSchemaGenerator();
        JSchema schema = generator.Generate(type, required != Required.Always);
        schema.Format = format;
        return schema;
    }
}

然后添加新生代提供者：

JSchemaGenerator generator = new JSchemaGenerator();
generator.GenerationProviders.Add(new FormatSchemaProvider());
JSchema schema = generator.Generate(typeof(MyClass));