使用GSON处理随机生成和不一致的JSON字段/密钥名称

pio*_*ion 14 java json gson

我有以下JSON代码段:

{ "randomlygeneratedKeyname0" : "some-value",
  "randomlygeneratedKeyname1": {
       "randomlygeneratedKeyname2" : {
           "randomlygeneratedKeyname3": "some-value",
           "randomlygeneratedKeyname4": "some-value"
       },
       "randomlygeneratedKeyname5": {
           "randomlygeneratedKeyname6": "some-value",
           "randomlygeneratedKeyname7": "some-value"
       }
   }
}
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请注意,我不知道它们的名称randomlygeneratedKeyname和它们的命名约定是不一致的,因此我无法创建相应的Java字段/变量名称.

我如何(在)GSON中序列化它?

在此先感谢您的帮助.

Jes*_*son 21

我很高兴地报告,GSON 2.0不费吹灰之力就支持默认地图和列表.像这样反序列化:

Object o = new Gson().fromJson(json, Object.class);
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其结果将是一个Map具有String键和任一StringMap值.

将该映射序列化为JSON,如下所示:

String json = new Gson().toJson(o);
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我们希望在2012年10月发布GSON 2.0.您可以从GSON SVN早期获得它.


Pro*_*uce 11

代码转储解决方案

import java.io.FileReader;
import java.lang.reflect.Type;
import java.util.HashMap;
import java.util.Map;
import java.util.Map.Entry;

import com.google.gson.Gson;
import com.google.gson.GsonBuilder;
import com.google.gson.JsonDeserializationContext;
import com.google.gson.JsonDeserializer;
import com.google.gson.JsonElement;
import com.google.gson.JsonObject;
import com.google.gson.JsonParseException;
import com.google.gson.reflect.TypeToken;

public class Foo
{
  public static void main(String[] args) throws Exception
  {
    Type mapStringObjectType = new TypeToken<Map<String, Object>>() {}.getType();

    GsonBuilder gsonBuilder = new GsonBuilder();
    gsonBuilder.registerTypeAdapter(mapStringObjectType, new RandomMapKeysAdapter());
    Gson gson = gsonBuilder.create();

    Map<String, Object> map = gson.fromJson(new FileReader("input.json"), mapStringObjectType);
    System.out.println(map);
  }
}

class RandomMapKeysAdapter implements JsonDeserializer<Map<String, Object>>
{
  @Override
  public Map<String, Object> deserialize(JsonElement json, Type unused, JsonDeserializationContext context)
      throws JsonParseException
  {
    // if not handling primitives, nulls and arrays, then just 
    if (!json.isJsonObject()) throw new JsonParseException("some meaningful message");

    Map<String, Object> result = new HashMap<String, Object> ();
    JsonObject jsonObject = json.getAsJsonObject();
    for (Entry<String, JsonElement> entry : jsonObject.entrySet())
    {
      String key = entry.getKey();
      JsonElement element = entry.getValue();
      if (element.isJsonPrimitive())
      {
        result.put(key, element.getAsString());
      }
      else if (element.isJsonObject())
      {
        result.put(key, context.deserialize(element, unused));
      }
      // if not handling nulls and arrays
      else
      {
        throw new JsonParseException("some meaningful message");
      }
    }
    return result;
  }
}
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  • 我忘了提一下,这个混乱与杰克逊只是一条线.Map map = new ObjectMapper().readValue(new File("input.json"),Map.class); (2认同)