Question

我是Swagger的新手。我使用Swagger编辑器编写了以下代码，但我收到了多个错误。我已经查看了文档和其他示例，但我无法澄清问题。

swagger: '2.0'
info:
  version: "1"
  title: 'Grocery Shop'
  description: 'This contains the url to access all the Grocery food related information. You can also add, update or delete any food item as well based on their respective id.'
  contact:
    email: you@your-company.com

tags:
- name: FoodItem
  description: Specifically for Food item
- name: Category
  description: Specifically for the category purpose

schemes:
 - https
paths:
  '/findfood/{foodItemId}':
    get:
      tags:
      - FoodItem
      summary: "Find the required food"
      description: "Find the food with passed food id as an argument. Make sure that passed id exist otherwise it will return an error"
      operationId: findFoodWithId
      produces:
      - "application/json"
      parameters:
      - in: path
        name: foodItemId
        required: true
        type: integer
        miniumum: 1
        description: The id associated with the food in the grocery shop

      responses:
        200:
          description: "successful operation"
          schema:
            $ref: '#definitions/Food'

        404:
          description: "Food is not found

definitions:
 Food:
    type: "object"

    properties:
     foodItemId:
      type: integer
      format: int32

    required:
    - foodItemId

    xml:
      name: "User"


# Added by API Auto Mocking Plugin
host: "localhost:8080"
basePath: "/grocery"

我得到的错误是：

paths./findfood/{foodItemId的语义错误}声明的路径参数   ＆＃34; foodItemId＆＃34;需要定义为路径参数   路径或操作级别跳转到第18行

paths./findfood/{foodItemId}.get.parameters[0]上的模式错误不是   一个来自   ＆LT;＃/定义/参数＆GT;，＆LT;＃/定义/ jsonReference＆GT;跳到线上   28

分析器错误无法读取块映射条目;多线键可以   不是隐含的键跳转到第48行

Answer 1

这些错误具有误导性。实际错误是：

miniumum: 1包含一个错字 - 迷你你妈妈
$ref: '#definitions/Food'在/和#之间需要definitions：'#/definitions/Food'
description: "Food is not found需要结束引号。

Swagger编辑器中paths./findfood/{foodItemId}的语义错误

1 个答案: