Swagger 创建 API 文档:Swagger Editor [英] Swagger create API document: Swagger Editor

问题描述

我正在使用 swagger 来记录我的 REST API 服务.我有我提供给服务的特定输入.我正在使用 swagger 编辑器自己创建 YAML 代码.我面临的问题是我无法将输入类型获取为 XML，默认情况下它采用 JSON.我的 yaml 代码中是否有任何问题.代码如下:

I am using swagger for documenting my REST API service. I have an specific input that I provide to the service. I am creating YAML code by myself using swagger editor. The issue I am facing is I am not able to get the input type as XML, it by default takes JSON. Is there any issue in my yaml code. The code is given below:

swagger: "2.0"
info:
title: Order Update to Dealers
 description: API description in Markdown.
 version: 1.0.0
host: #Host name cannot be specified here
basePath: /api/OrderUpdate
schemes:
  - http
paths:
/GetFullOrderAcknowlegement:
post:
  summary: Returns a list of users.
  consumes: 
    - application/xml 
  produces:
    - text/plain

  parameters:
    - in: body
      name: DealerInput
      description: Optional extended description in Markdown.

      schema:
        properties:
          DealerID:
            type: string
          PONumber:
            type: string
  responses:
    201:
      description: Created
    200:
      schema: {}
      description: OK
    401:
      schema: {}
      description: Authorization information is missing or invalid.

推荐答案

This是 Swagger Editor 3.3.0 和 Swagger UI 3.11.0 中的一个错误.已在 Editor 3.3.1 和 UI 3.12.0(2018 年 3 月 4 日发布)中修复.

This was a bug in Swagger Editor 3.3.0 and Swagger UI 3.11.0. It was fixed in Editor 3.3.1 and UI 3.12.0 (released on March 4, 2018).

作为一种解决方法，您可以下载编辑器 v3.2.9 并通过在浏览器中打开 index.html 文件在本地运行它.

As a workaround, you can download Editor v3.2.9 and run it locally by opening the index.html file in your browser.

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