如何在Swagger规范中接收动态响应 [英] How to receive a dynamic response in a Swagger spec

问题描述

我想通过我的API从数据库中请求一个表.但是，我不知道该表将具有多少列，或将包含什么列.如何在Swagger中指定呢?这就是我想做的:

I want to request a table from my database through my API. However, I don't know what number of columns the table will have, or what it will contain. How can I specify this in Swagger? This is what I would like to do:

paths:
  /reports/{id}:
    get:
      summary: Detailed results
      description: filler
      parameters:
        - name: id
          in: path
          description: filler
          required: true
          type: integer
          format: int64
      responses:
        200:
          description: OK
          schema:
            type: array
            items: 
              $ref: '#/definitions/DynamicObject'
definitions:
  DynamicObject:
    type: object
    properties:
      **$IDONTKNOWWHATTODO**

关于如何定义没有特定参数的JSON对象的任何想法?

Any ideas on how to define a JSON object with no specific parameters?

推荐答案

要描述任意JSON，请使用"type": "object".这是JSON中的示例:

To describe arbitrary JSON, please use "type": "object". Here is an example in JSON:

    "responses": {
      "200": {
        "description": "successful operation",
        "schema": {
          "type": "object"
        }
      }
    },

这篇关于如何在Swagger规范中接收动态响应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！