本文介绍了如何从OpenAPI 3.0 Yaml文件生成JSON示例?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我有我的openapi:3.0.0 YAML文件,我正在寻找一种从架构生成测试数据响应(JSON对象)的方法.这是我想要的,但是我无法在openapi:3.0.0上使用它,但代码有效非常适合摇摇欲坠":"2.0"的定义.我试图使代码与支持OpenAPI 3.0的Swagger Java库2.x一起使用.我知道我需要使用Swagger的2.x版本.

I have my openapi: 3.0.0 YAML file, I'm looking for a way to generate test data response (JSON object) from schema.This is what I am looking for, but I can't get it working for openapi: 3.0.0, the code works perfectly for "swagger": "2.0" definitions.I have tried to get the code working with Swagger Java libraries 2.x, which support OpenAPI 3.0. I know I need to use version 2.x of Swagger.

import io.swagger.parser.SwaggerParser;
import io.swagger.models.*;
import io.swagger.inflector.examples.*;
import io.swagger.inflector.examples.models.Example;
import io.swagger.inflector.processors.JsonNodeExampleSerializer;
import io.swagger.util.Json;
import io.swagger.util.Yaml;
import java.util.Map;
import com.fasterxml.jackson.databind.module.SimpleModule;


// Load your OpenAPI/Swagger definition
Swagger swagger = new SwaggerParser().read("http://petstore.swagger.io/v2/swagger.json");

// Create an Example object for the Pet model
Map<String, Model> definitions = swagger.getDefinitions();
Model pet = definitions.get("Pet");
Example example = ExampleBuilder.fromModel("Pet", pet, definitions, new HashSet<String>());
// Another way:
// Example example = ExampleBuilder.fromProperty(new RefProperty("Pet"), swagger.getDefinitions());

// Configure example serializers
SimpleModule simpleModule = new SimpleModule().addSerializer(new JsonNodeExampleSerializer());
Json.mapper().registerModule(simpleModule);

// Convert the Example object to string

// JSON example
String jsonExample = Json.pretty(example);
System.out.println(jsonExample);

此代码有效,只需为openapi获得相同的代码即可:3.0.0.

This code is working, just need to get the same code working for openapi: 3.0.0.

推荐答案

找到了解决方案

OpenAPI swagger = new OpenAPIV3Parser().read("url to Open API 3.0 Swagger")
Map < String, Schema > definitions = swagger.getComponents().getSchemas()
Schema model = definitions.get("Pet")
Example example = ExampleBuilder.fromSchema(model, definitions)
SimpleModule simpleModule = new SimpleModule().addSerializer(new JsonNodeExampleSerializer())
Json.mapper().registerModule(simpleModule)
String jsonExample = Json.pretty(example);
System.out.println(jsonExample);

这篇关于如何从OpenAPI 3.0 Yaml文件生成JSON示例?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!

09-11 12:23