如何在Swagger中指定GET参数的示例？

Question

我正在使用在线Swagger编辑器为我的API创建Swagger规范.

我的API有一个GET请求端点,我使用以下YAML代码来描述输入参数:

paths:
  /fooBar:
    get:
      tags:
        - foobar
      summary: ''
      description: ''
      operationId: foobar
      consumes:
        - application/x-www-form-urlencoded
      produces:
        - application/json
      parameters:
        - name: address
          in: query
          description: Address to be foobared
          required: true
          type: string
          example: 123, FakeStreet
        - name: city
          in: query
          description: City of the Address
          required: true
          type: string
          example: New York

如果我放入example标签,我会收到错误消息:

不是<#/ definitions/parameter>,<#/ definitions/jsonReference>中的一个

如何在Swagger中编写GET参数时设置示例？

Answer 1

OpenAPI/Swagger 2.0没有example非body参数的关键字.您可以在参数中指定示例description.一些工具,如Swagger UI v2,v3.12 +和Dredd也x-example为此目的支持扩展属性:

      parameters:
        - name: address
          in: query
          description: Address to be foobared. Example: `123, FakeStreet`.  # <-----
          required: true
          type: string
          x-example: 123, FakeStreet   # <-----

OpenAPI 3.0本身支持参数示例:

      parameters:
        - name: address
          in: query
          description: Address to be foobared
          required: true
          schema:
            type: string
            example: 123, FakeStreet   # <----
          example: 456, AnotherStreet  # Overrides schema-level example