如何从示例JSON生成Swagger #definition [英] How to generate a Swagger #definition from sample JSON
问题描述
从宠物商店示例中获取以下#definition.给定#definition部分,可以生成JSON结构
Take the following #definition from the pet store example. Given a #definition section a JSON structure can be generated
例如
在复杂的JSON文件较大的情况下,是否可以进行相反的操作?
Is there something that can do the reverse given a largeish complex JSON file?
鉴于下面的JSON结构,我可以获取为保存某些输入而生成的swagger文件的#defintion部分
Given the below JSON Structure can I get the #defintion section of a swagger file generated to save some typing
{
"variable": "sample",
"object1": {
"obj-field1": "field 1 of object",
"obj-field2": "field 2 of object",
"anArray": [
"Value 1",
{
"anArrayObj1": "obj1fieldinarray",
"anArrayObj2": "obj2fieldinarray"
}
]
}
}
推荐答案
您可以使用以下JSON到OpenAPI的模式转换器:
https://roger13.github.io/SwagDefGen/
You can use this JSON-to-OpenAPI schema converter:
https://roger13.github.io/SwagDefGen/
( GitHub项目)
虽然我还没有亲自使用它,所以我不确定它有多好.
I haven't used it personally though, so I'm not sure how good it is.
由于OpenAPI使用JSON模式的子集,因此您也可以使用 JSON模式生成器之一,但是您可能需要手动调整生成的定义以使其与OpenAPI兼容.
Since OpenAPI uses a subset of JSON Schema, you could also use one of the JSON Schema generators, however you may need to manually tweak the generated definition to make it OpenAPI-compatible.
这篇关于如何从示例JSON生成Swagger #definition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!