[英]Swagger Annotation - not to include parameter when there is no parameter?
I have swagger-jersey2-jaxrs 1.5.1-M2 integrated with Drop-wizard. 我有与Drop-wizard集成的swagger-jersey2-jaxrs 1.5.1-M2。 I have generated the following yaml with the swagger annotations I have included on my resource and models:
我使用资源和模型中包含的大头贴注释生成了以下yaml:
---
swagger: "2.0"
info:
version: "1.0.0"
title: "test Application"
tags:
- name: "test"
paths:
/v1/test/version:
get:
tags:
- "v1test"
summary: "version number of test"
description: "Returns version number of test"
operationId: "getVersionAndBuildInfo"
produces:
- "application/json"
parameters:
responses:
404:
description: "build number not found."
200:
description: "successful operation"
schema:
type: "array"
items:
$ref: "#/definitions/testInformation"
definitions:
testInformation:
properties:
build:
type: "string"
description: "test build number"
version:
type: "string"
description: "test version number"
buildTimestamp:
type: "string"
description: "test build timestamp"
description: "model for test version information"
notice that parameters: is include in the generated yaml from swagger. 注意, 参数:是包含在从摇号生成的yaml中的。 when I paste this in the swagger editor 2.0 it is throwing an error saying "Array is too short (0), minimum 1".
当我将其粘贴到swagger编辑器2.0中时,它抛出一个错误,提示“数组太短(0),最小值为1”。
Is there a way not to generate the parameters: in yaml when the URL does not require any parameters? 有没有一种方法可以不生成参数:在yaml中,如果URL不需要任何参数? I didn't include any parameter annotation and it still shows up on the generated yaml.
我没有包含任何参数注释,它仍然显示在生成的Yaml中。
We had a bug in the JSON Schema for Swagger definitions that didn't allow for an empty parameters. Swagger定义的JSON模式中存在一个错误,该错误不允许使用空参数。 This was fixed, and is going to be available in swagger-editor as well soon.
此问题已修复,很快将在swagger-editor中可用。
Please follow https://github.com/swagger-api/swagger-editor/issues/395 to see when it is updated. 请遵循https://github.com/swagger-api/swagger-editor/issues/395查看更新时间。
There's no point trying to 'solve' this issue in the generation process because the output is not wrong. 没有必要尝试在生成过程中“解决”此问题,因为输出没有错误。
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