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构建模糊器.

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

我想测试来自OpenAPI定义的所有可用路径，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，给定此模型:

...
"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类似以下内容:

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偏转器库具有 ExampleBuilder 类正是为此目的而设计的.它使您可以从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.

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);

上面的示例使用了Swagger Java库1.x，该库支持OpenAPI 2.0定义(swagger: '2.0').

The example above uses Swagger Java libraries 1.x, which support OpenAPI 2.0 definitions (swagger: '2.0').

如果您的API定义是OpenAPI 3.0(openapi: 3.0.0)，则需要使用Swagger Java库的2.x版本，并相应地更新导入和类名，例如io.swagger.parser.SwaggerParser→ io.swagger.v3.parser.OpenAPIV3Parser等

If your API definition is OpenAPI 3.0 (openapi: 3.0.0), you need to use version 2.x of Swagger Java libraries, and update the imports and class names appropriately, e.g. io.swagger.parser.SwaggerParser → io.swagger.v3.parser.OpenAPIV3Parser, etc.

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