Swagger 如何在招摇规范中接收动态响应
我想通过API从数据库中请求一个表。但是,我不知道表将有多少列,或者它将包含什么。我如何在招摇过市中指定这一点?这就是我想做的:Swagger 如何在招摇规范中接收动态响应,swagger,swagger-2.0,openapi,Swagger,Swagger 2.0,Openapi,我想通过API从数据库中请求一个表。但是,我不知道表将有多少列,或者它将包含什么。我如何在招摇过市中指定这一点?这就是我想做的: paths: /reports/{id}: get: summary: Detailed results description: filler parameters: - name: id in: path description: filler
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对象有什么想法吗?要描述任意JSON,请使用
“type”:“object”
。以下是一个JSON示例:
"responses": {
"200": {
"description": "successful operation",
"schema": {
"type": "object"
}
}
},