Swagger 参数和复杂类型 [英] Swagger Parameters and Complex Types

问题描述

在以下 Swagger 定义中，我需要参数 labelValue 的类型为 LabelValueObject，以便对其进行验证并正确反序列化.但是，我无法弄清楚语法！怎么做?

In the following Swagger definition, I need the parameter labelValue to be of type LabelValueObject, so that it will be validated and correctly deserialized. However, I can't figure out the syntax! How can that be done?

swagger: "2.0"

paths:
  /competition:
    post:
      parameters:
        - name: labelValue
          in: formData
          type: array
          items:
            type: string       ### this has to be a LabelValueObject ###
      responses:
        default:
          description: Error
          schema:
            $ref: "#/definitions/AnyResponse"

definitions:
  AnyResponse:
    properties:
      any:
        type: string
  LabelValueObject:
    properties:
      label:
        type: string
      value:
        type: string
    required:
      - label
      - value

推荐答案

将对象作为参数传递的唯一方法是将其放入正文 (in: body) 然后定义这个schema 中的对象(使用 $ref 内联定义或引用预定义的对象).这是一个完整的示例:

The only way to pass an object as a parameter is to put it in the body (in: body) and then define this object in schema (inline definition or reference to an predefined object with $ref). Here's a full example:

swagger: "2.0"

info:
  title: A dummy title
  version: 1.0.0

paths:
  /competition:
    post:
      parameters:
        - name: labelValue
          in: body
          schema:
            $ref: '#/definitions/LabelValueObject'
      responses:
        default:
          description: Error
          schema:
            $ref: "#/definitions/AnyResponse"

definitions:
  AnyResponse:
    properties:
      any:
        type: string
  LabelValueObject:
    properties:
      label:
        type: string
      value:
        type: string
    required:
      - label
      - value

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