在我的"简化"API中,所有响应都是从基础"响应"类派生(继承)的.响应类由填充元数据的标头和包含用户请求的核心数据的主体组成.响应(在JSON中)的布局使得所有元数据都在第一个"层"上,而主体是一个称为"body"的单个属性,因此
response
|--metadata attribute 1 (string/int/object)
|--metadata attribute 2 (string/int/object)
|--body (object)
|--body attribute 1 (string/int/object)
|--body attribute 2 (string/int/object)
Run Code Online (Sandbox Code Playgroud)
我试图使用以下JSON在swagger中定义这种关系:
{
...
"definitions": {
"response": {
"allOf": [
{
"$ref": "#/definitions/response_header"
},
{
"properties": {
"body": {
"description": "The body of the response (not metadata)",
"schema": {
"$ref": "#/definitions/response_body"
}
}
}
}
]
},
"response_header": {
"type": "object",
"required": [
"result"
],
"properties": {
"result": {
"type": "string",
"description": "value of 'success', for a successful response, or 'error' if there is an error",
"enum": [
"error",
"success"
]
},
"message": {
"type": "string",
"description": "A suitable error message if something went wrong."
}
}
},
"response_body": {
"type": "object"
}
}
}
Run Code Online (Sandbox Code Playgroud)
然后我尝试通过创建从body/header继承的各种body/header类来创建不同的响应,然后创建由相关的header/body类组成的子响应类(在底部的源代码中显示).但是,我确信这是做错事的方法,或者我的实现是不正确的.我一直无法在swagger 2.0规范中找到继承的例子(如下所示),但已经找到了一个组合的例子.
我很确定这个"鉴别器"有很大的作用,但不确定我需要做什么.
有人可以告诉我如何在swagger 2.0(JSON)中实现组合+继承,最好通过"修复"下面的示例代码.如果我可以指定一个继承自响应的ErrorResponse类,其中标题中的"result"属性始终设置为"error",那也是很好的.
{
"swagger": "2.0",
"info": {
"title": "Test API",
"description": "Request data from the system.",
"version": "1.0.0"
},
"host": "xxx.xxx.com",
"schemes": [
"https"
],
"basePath": "/",
"produces": [
"application/json"
],
"paths": {
"/request_filename": {
"post": {
"summary": "Request Filename",
"description": "Generates an appropriate filename for a given data request.",
"responses": {
"200": {
"description": "A JSON response with the generated filename",
"schema": {
"$ref": "#/definitions/filename_response"
}
}
}
}
}
},
"definitions": {
"response": {
"allOf": [
{
"$ref": "#/definitions/response_header"
},
{
"properties": {
"body": {
"description": "The body of the response (not metadata)",
"schema": {
"$ref": "#/definitions/response_body"
}
}
}
}
]
},
"response_header": {
"type": "object",
"required": [
"result"
],
"properties": {
"result": {
"type": "string",
"description": "value of 'success', for a successful response, or 'error' if there is an error",
"enum": [
"error",
"success"
]
},
"message": {
"type": "string",
"description": "A suitable error message if something went wrong."
}
}
},
"response_body": {
"type": "object"
},
"filename_response": {
"extends": "response",
"allOf": [
{
"$ref": "#definitions/response_header"
},
{
"properties": {
"body": {
"schema": {
"$ref": "#definitions/filename_response_body"
}
}
}
}
]
},
"filename_response_body": {
"extends": "#/definitions/response_body",
"properties": {
"filename": {
"type": "string",
"description": "The automatically generated filename"
}
}
}
}
}
Run Code Online (Sandbox Code Playgroud)
为了尝试澄清我想要的东西,我创建了下面的基本图表,其中旨在显示所有响应都是"响应"对象的实例化,这些对象是由(组合)使用response_header和response_body对象的任意组合构建的.response_header和response_body对象可以扩展并插入到任何响应对象中,这在filename_response的情况下完成,该filename_response使用基本response_body类的filename_response_body子节点.错误和成功响应都使用"响应"对象.
Tom*_*ski 103
作为一个招摇的初学者,我没有找到关于polimorphism和组成的官方文档很容易理解,因为它缺乏一个例子.当我搜索网络时,有很多很好的例子在extends
有效的时候引用了swagger 1.2 .
对于swagger 2.0,我通过这个谷歌小组在github上找到了一个很好的例子
基于以上来源,这是YAML中的一个简短的有效继承示例:
definitions:
Pet:
discriminator: petType
required:
- name
- petType # required for inheritance to work
properties:
name:
type: string
petType:
type: string
Cat:
allOf:
- $ref: '#/definitions/Pet' # Cat has all properties of a Pet
- properties: # extra properties only for cats
huntingSkill:
type: string
default: lazy
enum:
- lazy
- aggressive
Dog:
allOf:
- $ref: '#/definitions/Pet' # Dog has all properties of a Pet
- properties: # extra properties only for dogs
packSize:
description: The size of the pack the dog is from
type: integer
Run Code Online (Sandbox Code Playgroud)
obl*_*lex 20
我发现即使没有定义,组合也能正常工作discriminator
.
例如,基数Response
:
definitions:
Response:
description: Default API response
properties:
status:
description: Response status `success` or `error`
type: string
enum: ["success", "error"]
error_details:
description: Exception message if called
type: ["string", "object", "null"]
error_message:
description: Human readable error message
type: ["string", "null"]
result:
description: Result body
type: ["object", "null"]
timestamp:
description: UTC timestamp in ISO 8601 format
type: string
required:
- status
- timestamp
- error_details
- error_message
- result
Run Code Online (Sandbox Code Playgroud)
呈现为:
我们可以扩展它以改进result
字段的自定义模式:
FooServiceResponse:
description: Response for Foo service
allOf:
- $ref: '#/definitions/Response'
- properties:
result:
type: object
properties:
foo_field:
type: integer
format: int32
bar_field:
type: string
required:
- result
Run Code Online (Sandbox Code Playgroud)
它将被正确呈现为:
注意,这allOf
足以使其工作并且不discriminator
使用任何字段.这很好,因为它很有用,这很重要,因为我认为,工具将能够生成没有discriminator
字段的代码.
这里的所有答案都已经很好了,但我只想添加关于组合与继承的小注释.根据Swagger/OpenAPI规范,要实现组合,使用该allOf
属性就足够了,正如@oblalex正确指出的那样.但是,要实现继承,您需要使用allOf
with discriminator
,如@TomaszSętkowski中的示例.
此外,我在API Handyman中发现了一些更多Swagger 组合和继承的例子.他们是Arnaud Lauret 出色的Swagger/OpenAPI教程系列的一部分,我想每个人都应该看看.
归档时间: |
|
查看次数: |
61961 次 |
最近记录: |