我目前正在使用Swagger 2.0记录API。
API会生成application/json
和application/xml
。
定义部分是:
definitions:
newsFeed:
type: object
description: the object is wrapped...
properties:
feedId:
type: string
example: '987654'
xml:
attribute: true
title:
type: string
description:
type: string
feedUrl:
type: string
example: http://heise.de.feedsportal.com/c/35207/f/653902/index.rss
logoUrl:
type: string
example: http://example.com/logo.png
language:
type: string
example: de_DE
pos:
type: string
example: '0'
active:
type: boolean
Swagger-UI生成此XML。
<?xml version="1.0" encoding="UTF-8"?>
<newsFeed feedId="987654">
<title>string</title>
<description>string</description>
<feedUrl>http://heise.de.feedsportal.com/c/35207/f/653902/index.rss</feedUrl>
<logoUrl>http://example.com/logo.png</logoUrl>
<language>de_DE</language>
<pos>0</pos>
<active>true</active>
</newsFeed>
这个JSON:
{
"feedId": "987654",
"title": "string",
"description": "string",
"feedUrl": "http://heise.de.feedsportal.com/c/35207/f/653902/index.rss",
"logoUrl": "http://example.com/logo.png",
"language": "de_DE",
"pos": "0",
"active": true
}
两者都是正确的,但API提供了不同的JSON。
{
"newsFeed": {
"feedId": "987654",
"title": "string",
"description": "string",
"feedUrl": "http://heise.de.feedsportal.com/c/35207/f/653902/index.rss",
"logoUrl": "http://example.com/logo.png",
"language": "de_DE",
"pos": "0",
"active": true
}
}
如何使用Swagger记录此内容?