Flasgger:参数不显示和架构

Question

我最近开始学习Swagger for APi文档。然而，当我转到localhost:5000/apidocs时，我看到了我添加的API，但是所有的API都没有参数，即使我已经在yml文件中描述了参数。我在网上找不到任何相关的东西。

下面是我的YML代码：

paths:
  /api/ppp:
    get:
      tags:
      - "user"
      summary: "Find User by uID"
      description: "Returns a single user info"
      operationId: "getuserById"
      produces:
        - "application/xml"
        - "application/json"
      parameters:
        - name: "userId"
          in: "path"
          description: "ID of user to return"
          required: true
          type: "uuid"

下面是我如何将swagger添加到api @swag_from('pathto/swagger.yml')

同样，我还有另一个疑问。我们在yml规范文件中定义模式。我想知道这些模式到底代表了什么，有必要使用它们吗？

附件是APIdocs页面的屏幕截图

​

Answer 1

试试这个：

tags:
- user
summary: "Find User by uID"
description: "Returns a single user info"
operationId: "getuserById"
produces:
  - "application/xml"
  - "application/json"
parameters:
  - name: "userId"
    in: "path"
    description: "ID of user to return"
    required: true
    type: "uuid"