在Swagger中，如何定义使用文件以及模式参数的API? [英] In Swagger, how to define an API that consumes a file along with a schema parameter?

问题描述

我正在尝试使用Swagger定义一个接受实际文件的API和一个描述文件内容的架构对象.这是Swagger YAML的片段.但是，它不会在Swagger编辑器中验证.

I am trying to use Swagger to define an API that accepts an actual file and a schema object that describes the contents of a file. Here is a snippet of the Swagger YAML. However it won't validate in the Swagger Editor.

/document:
  post:
    summary: Api Summary
    description: Api Description
    consumes:
      - multipart/form-data
    parameters:
      - name: documentDetails
        in: formData
        description: Document Details
        required: true
        schema:
          $ref: '#/definitions/Document'
      - name: document
        in: formData
        description: The actual document
        required: true
        type: file

Swagger编辑器会引发以下验证错误:

The Swagger Editor throws the following validation error:

挥舞着错误:数据与"oneOf"中的任何架构都不匹配

Swagger Error: Data does not match any schemas from 'oneOf'

我错过了什么吗?还是这不是Swagger支持的功能?

Am I missing something? Or Is this not a supported feature of Swagger?

推荐答案

swagger不支持formData中的'object'类型，仅作为主体参数.

swagger does not support type 'object' in formData, only as body parameters.

这篇关于在Swagger中，如何定义使用文件以及模式参数的API?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！