如何通过通用机制让Swagger UI显示简化响应体（无需添加属性）

我有一个API端点，期望Swagger UI中localDate字段的响应示例格式如下：

{
"localDate": "Wednesday, 13 March 2024"
}

但在Swagger UI中调用该端点后，实际显示的响应是LocalDate的完整对象结构：

{
  "localDate": {
    "calendar": {
      "id": "ISO",
      "name": "ISO",
      "minYear": -9998,
      "maxYear": 9999,
      "eras": [
        {
          "name": "BCE"
        },
        {
          "name": "CE"
        }
      ]
    },
    "year": 2024,
    "month": 3,
    "day": 14,
    "dayOfWeek": 4,
    "yearOfEra": 2024,
    "era": {
      "name": "CE"
    },
    "dayOfYear": 74
  }
}

相关代码 #

端点代码 #

[HttpGet("any-simple-random-object")]
[Produces("application/json")]
[ProducesResponseType(StatusCodes.Status200OK)]
public async Task<ActionResult<AnySimpleObjectResponse>> GetAnySimpleObject()
=> await Task.FromResult(AnySimpleObjectResponse.CreateRandomObject());

响应生成器代码 #

public record AnySimpleObjectResponse(LocalDate LocalDate)
{
    public static AnySimpleObjectResponse CreateRandomObject()
    => new(LocalDate.FromDateTime(DateTime.Now));
}

需求 #

我需要实现：

• Swagger UI中显示的响应体与预期示例一致
• Postman等外部工具调用时，仍能获取LocalDate的完整对象结构
• 采用通用机制，无需为每个响应对象或端点单独添加属性

内容的提问来源于stack exchange，提问作者MarekBe