如何包装JSON对象? [英] How do I wrap JSON objects?

问题描述

我目前正在寻找一种在Swagger UI组件中包装JSON的方法.

I'm currently looking for a way to wrap JSON in the Swagger UI component.

在YAML中，我的对象声明是:

In YAML my object declaration is:

  restException:
    properties:
      message:
        type: string

Swagger UI生成的输出是(我同意，这是正确的): { "message": "string" }

The output generated by Swagger UI is (whic I agree, is correct): { "message": "string" }

我想要的是:

"restException": {
  "message": "string"
}

通过在YAML文件中显式声明包装器，我找到了一种丑陋的方法.但这很不好，因为当我使用"Swagger Codegen"生成客户端或服务器代码时也会生成它.

I've just find a ugly way to do it by explicitely declaring the wrapper in the YAML file. But it's verry bad since it's also generates when I use "Swagger Codegen" to generate client or server code.

restExceptionContainer: restException: properties: message: type: string

restExceptionContainer: restException: properties: message: type: string

如果可以的话，我可以在Swagger UI文件中添加代码！需要您的帮助以找到位置:)

I'm ok for adding code in the Swagger UI file if needed ! Need your help to find where :)

推荐答案

您应将restException文档记录为对象(类型:object).

You should document restException as an object (type: object).

请参考

Please refer to https://github.com/swagger-api/swagger-codegen/blob/master/modules/swagger-codegen/src/test/resources/2_0/petstore.yaml#L646 as an example and look at how Pet and Category are defined.

  Pet:
    type: object
    required:
      - name
      - photoUrls
    properties:
      id:
        type: integer
        format: int64
      category:
        $ref: '#/definitions/Category'

其中类别"定义为:

  Category:
    type: object
    properties:
      id:
        type: integer
        format: int64
      name:
        type: string

这篇关于如何包装JSON对象?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！