如何从 OpenAPI/Swagger 模型定义生成 JSON 示例? [英] How to generate JSON examples from OpenAPI/Swagger model definition?

问题描述

我正在为具有 OpenAPI (Swagger) 定义的 REST API 构建一个模糊器.

I'm building a fuzzer for a REST API that has an OpenAPI (Swagger) definition.

我想测试 OpenAPI 定义中的所有可用路径，生成数据以测试服务器，分析响应代码和内容，并验证响应是否符合 API 定义.

I want to test all available path from the OpenAPI definition, generate data to test the servers, analyse responses code and content, and to verify if the responses are conform to the API definition.

我正在寻找一种从模型定义中生成数据(JSON 对象)的方法.

I'm looking for a way to generate data (JSON object) from model definitions.

例如，给定这个模型:

...
"Pet": {
  "type": "object",
  "required": [
    "name",
    "photoUrls"
  ],
  "properties": {
    "id": {
      "type": "integer",
      "format": "int64"
    },
    "category": {
      "$ref": "#/definitions/Category"
    },
    "name": {
      "type": "string",
      "example": "doggie"
    },
    "photoUrls": {
      "type": "array",
      "items": {
        "type": "string"
      }
    },
    "tags": {
      "type": "array",
      "items": {
        "$ref": "#/definitions/Tag"
      }
    },
    "status": {
      "type": "string",
      "description": "pet status in the store"
    }
  }
}

我想生成随机数据并得到这样的东西:

I want to generate random data and get something like this:

{
  "id": 0,
  "category": {
    "id": 0,
    "name": "string"
  },
  "name": "doggie",
  "photoUrls": [
    "string"
  ],
  "tags": [
    {
      "id": 0,
      "name": "string"
    }
  ],
  "status": "string"
}

推荐答案

Swagger Inflector 库有 ExampleBuilder 类正是为此目的.它允许您从 OpenAPI (Swagger) 定义中的模型生成 JSON、XML 和 YAML 示例.

The Swagger Inflector library has the ExampleBuilder class exactly for this purpose. It lets you generate JSON, XML and YAML examples from models in an OpenAPI (Swagger) definition.

要使用 OpenAPI 2.0 (swagger: '2.0') 定义，请使用 Swagger Java 库 1.x.

To work with OpenAPI 2.0 (swagger: '2.0') definitions, use Swagger Java libraries 1.x.

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);
Yaml.mapper().registerModule(simpleModule);

// Convert the Example object to string

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

// YAML example
String yamlExample = Yaml.pretty().writeValueAsString(example);
System.out.println(yamlExample);

// XML example (TODO: pretty-print it)
String xmlExample = new XmlExampleSerializer().serialize(example);
System.out.println(xmlExample);

OpenAPI 3.0 示例

对于 OpenAPI 3.0 示例，查看此答案.您需要 Swagger Java 库的 2.x 版本，并适当地更新导入和类名，例如将 io.swagger.parser.SwaggerParser 更改为 io.swagger.v3.parser.OpenAPIV3Parser 等等.

OpenAPI 3.0 example

For an OpenAPI 3.0 example, see this answer. You need version 2.x of Swagger Java libraries, and update the imports and class names appropriately, e.g. change io.swagger.parser.SwaggerParser to io.swagger.v3.parser.OpenAPIV3Parser and so on.

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