Question

[
  {
      "updated_at":"2012-03-02 21:06:01",
      "fetched_at":"2012-03-02 21:28:37.728840",
      "description":null,
      "language":null,
      "title":"JOHN",
      "url":"http://rus.JOHN.JOHN/rss.php",
      "icon_url":null,
      "logo_url":null,
      "id":"4f4791da203d0c2d76000035",
      "modified":"2012-03-02 23:28:58.840076"
   },
   {
      "updated_at":"2012-03-02 14:07:44",
      "fetched_at":"2012-03-02 21:28:37.033108",
      "description":null,
      "language":null,
      "title":"PETER",
      "url":"http://PETER.PETER.lv/rss.php",
      "icon_url":null,
      "logo_url":null,
      "id":"4f476f61203d0c2d89000253",
      "modified":"2012-03-02 23:28:57.928001"
   }
]

Answer 1

首先为你的json数据创建一个类（Json_obj）。然后你可以试试这个：

String json_str='[ { "updated_at":"2012-03-02 21:06:01", "fetched_at":"2012-03-02 21:28:37.728840", "description":null, "language":null, "title":"JOHN", "url":"http://rus.JOHN.JOHN/rss.php", "icon_url":null, "logo_url":null, "id":"4f4791da203d0c2d76000035", "modified":"2012-03-02 23:28:58.840076" }, { "updated_at":"2012-03-02 14:07:44", "fetched_at":"2012-03-02 21:28:37.033108", "description":null, "language":null, "title":"PETER", "url":"http://PETER.PETER.lv/rss.php", "icon_url":null, "logo_url":null, "id":"4f476f61203d0c2d89000253", "modified":"2012-03-02 23:28:57.928001" } ]';
Gson gson = new Gson();
Json_obj json_obj = gson.fromJson(json_str, Json_obj.class);

现在你的json数据在一个对象中转换。你可以从该对象中获取任何值。

Answer 2

我建议您只使用jackson库。

你可以快点

ObjectMapper mapper = new ObjectMapper () // this reads json to Pojo and writes Pojo to json

YourPojoClass obj = mapper.readValue (....)

参考：mkyong.com/java/jackson-2-convert-java-object-to-from-json

如何用以下结构解析Json

2 个答案: