首页
学习
活动
专区
圈层
工具
发布
社区首页 >问答首页 >如何在Swagger中添加来自组件的标题?

如何在Swagger中添加来自组件的标题?
EN

Stack Overflow用户
提问于 2022-10-27 14:57:42
回答 1查看 42关注 0票数 0
代码语言:javascript
复制
        post:
          tags:
            - Customer
          summary: Create Customer
          description: Create Customer
          operationId: Customer
          parameters:
            - in: header
              name: operationId
              description: Content Type
              required: true
              schema:
                type: string
                example: Customer
          requestBody:
              $ref: '#/components/requestBodies/createCustomer'      
          responses:
           '200':
              $ref: '#/components/responses/postsuccess'
           '400':
              $ref: '#/components/responses/postfailed'   

components:
  headers:
    Content-Type:
      description: request content type
      required: true
      schema:
        type: string
        example: application/json

如何在Swagger中添加来自组件的标题?

EN

回答 1

Stack Overflow用户

发布于 2022-10-27 17:17:58

在OpenAPI中,Content-Type是一个特殊的标头,不能将其描述为标头参数。相反,它是使用请求/响应媒体类型定义的。

例如,POST请求主体:

代码语言:javascript
复制
post:
  requestBody:
    required: true
    content:
      application/json:  # <--- This line defines the "Content-Type" header in requests
        schema:
          $ref: '#/components/schemas/MyObject'

或者当$ref从requestBody中删除components时(如您的示例中所示),您将得到如下内容:

代码语言:javascript
复制
post:
  requestBody:
    $ref: '#/components/requestBodies/createCustomer'

...

components:
  requestBodies:
    createCustomer:
      required: true
      content:
        application/json:   # <--- This defines the "Content-Type" header in requests
          schema:
            $ref: '#/components/schemas/Customer'
票数 0
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/74224165

复制
相关文章

相似问题

领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档