How do I wrap JSON objects?

问题

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

In YAML my object declaration is:

  restException:
    properties:
      message:
        type: string

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

and what I want is:

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

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

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

回答1:

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

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'

where Category is defined as:

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

来源：https://stackoverflow.com/questions/36545405/how-do-i-wrap-json-objects