如何重用swagger定义并删除其中的一些参数? [英] How to reuse swagger definitions and remove some of the parameters in it?

问题描述

这是我的代码:

definitions:
  User:
    type: object
    properties:
      id:
        type: integer
      username:
        type: string
      first_name:
        type: string
      last_name:
        type: string
      password:
        type: string
      created_at:
        type: string
        format: date-time
      updated_at:
        type: string
        format: date-time
    required:
      - username
      - first_name
      - last_name
      - password

/api/users:
  post:
    description: Add a new user
    operationId: store
    parameters:
      - name: user
        description: User object
        in: body
        required: true
        type: string
        schema:
          $ref: '#/definitions/User'
    produces:
      - application/json
    responses:
      "200":
        description: Success
        properties:
          success:
            type: boolean
          data:
            $ref: '#/definitions/User'

如您所见，在 /api/users 下的 post 键中，我使用了 User 定义作为我的架构.

As you can see, in the post key under /api/users I used the User definition as my schema on it.

我想减少我的代码，所以我重用了 User 定义作为我的模式.这里的问题是我不需要 id、created_at 和 updated_at 字段.

I want to lessen my code so I reused the User definition as my schema. The problem here is that I do not need the id, created_at and updated_at fields.

除了提到的字段之外，有没有办法只继承一些字段?另外，由于我正在尝试学习 swagger，因此我希望有一些建议可以让它变得更好.谢谢.

Is there a way to just inherit some of the fields except the fields mentioned? Also, I would love some suggestions to make it better since I'm trying to learn swagger. Thank you.

推荐答案

如 此答案 类似问题:

您必须单独定义模型.

但是，您可以选择排除和差异的情况.

However, you have options for the cases of exclusion and difference.

如果您想排除，这是简单的情况，请创建一个模型使用排除的属性，比如 ModelA.然后将 ModelB 定义为ModelA 加上附加属性:

If you're looking to exclude, which is the easy case, create a model of with the excluded property, say ModelA. Then define ModelB as ModelA plus the additional property:

ModelB:
  allOf:
    - $ref: "#/definitions/ModelA"
    - type: object
      properties:
        id:
          type: string

如果您想定义差异，请遵循相同的方法上面，并从 ModelA 中排除 id.然后定义ModelB和ModelC作为扩展 ModelA 并向它们添加 id 属性，每个属性都有自己的限制.请注意，JSON Schema 可以让您遵循上面的原始示例在某些情况下覆盖"了定义.但是，由于它并不是真正的压倒一切，因此需要更好地理解 JSON Schema 的概念，不要简单化错误，我建议现在走这条路.

If you're looking to define the difference, follow the same method above, and exclude the id from ModelA. Then define ModelB and ModelC as extending ModelA and add the id property to them, each with its own restrictions. Mind you, JSON Schema can allow you to follow the original example above for some cases to "override" a definition. However, since it is not really overriding, and one needs to understand the concepts of JSON Schema better to not make simple mistakes, I'd recommend going this path for now.

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