Question

我有一个javax.json.Json对象，我需要验证其是否为有效的Swagger文件。我写了这些实用程序功能-

package com.somecompany.gis.util;

import java.io.IOException;
import java.io.StringWriter;
import java.util.Map;

import javax.json.Json;
import javax.json.JsonObject;
import javax.json.JsonWriter;

import com.fasterxml.jackson.databind.JsonNode;
import com.fasterxml.jackson.databind.ObjectMapper;

import io.swagger.models.Swagger;
import io.swagger.parser.SwaggerParser;



public class Converter {
public static JsonNode toJsonNode(JsonObject jsonObject) throws IOException {

    // Parse a JsonObject into a JSON string
    StringWriter stringWriter = new StringWriter();
    try (JsonWriter jsonWriter = Json.createWriter(stringWriter)) {
        jsonWriter.writeObject(jsonObject);
    }
    String json = stringWriter.toString();

    // Parse a JSON string into a JsonNode
    ObjectMapper objectMapper = new ObjectMapper();
    JsonNode jsonNode = objectMapper.readTree(json);

    return jsonNode;
}

    public static boolean isValidSwaggerSpec(JsonObject jsonObject) {
    try {
        JsonNode jsonNode = toJsonNode(jsonObject);
        Swagger swagger = new SwaggerParser().read(jsonNode);
        return true;
    }catch(IOException ioe) {
        return false;
    }catch(Exception e) {
        return false;
    }

}

但是，我看到即使使用无效的Swagger文件，我也得到了true的评估。有什么方法可以检查Swagger是否有效？

Answer 1

您也许可以使用SwaggerParser#readWithInfo返回一个SwaggerDeserializationResult对象，当发生错误时，它们会用特定的消息填充该对象的List<String> messages：

return new SwaggerDeserializationResult().message("empty or null swagger supplied");
result = new SwaggerDeserializationResult().message("Definition does not appear to be a valid Swagger format");
return new SwaggerDeserializationResult().message("malformed or unreadable swagger supplied");

您可以检查这些句子并确保它们不存在，这意味着它已成功解析。

SwaggerDeserializationResult swagger = new SwaggerParser().readWithInfo(myJSONString);
List<String> messages = swagger.getMessages();

// Check if messages contains any of those strings

检查JSON是否是有效的Swagger

1 个答案: