如何解析具有相同结构但名称不同的json数组

我想解析一个json文件，但它是这样的：

CDG: { id: "32569", airport_name: "Charles De Gaulle", latitude: "49.0167", longitude: "2.55", timezone: "2", dst_indicator: "E", city: "Paris", country: "France", country_code: "FR", region: "TC1", listing_display: "true", pseudonyms: "" }, ORY: { id: "33539", airport_name: "Orly", latitude: "48.7167", longitude: "2.3833", timezone: "2", dst_indicator: "E", city: "Paris", country: "France", country_code: "FR", region: "TC1", listing_display: "true", pseudonyms: "" }, LBG: { id: "123425", airport_name: "Le Bourget", latitude: "48.969444", longitude: "2.441389", timezone: "1", dst_indicator: "E", city: "Paris", country: "France", country_code: "FR", region: "TC1", listing_display: "true", pseudonyms: "" }, 

但是大约有三千个像这样的对象。 我一直在使用Gson来解析我的json对象，但是我如何解析这种文件呢？ 我怎样才能找到名字“CDG”或“ORY”？

Answer 1

你可以尝试这样的事情：

  String str = "{CDG: {\n" + "id: \"32569\",\n" + "airport_name: \"Charles De Gaulle\",\n" + "latitude: \"49.0167\",\n" + "longitude: \"2.55\",\n" + "timezone: \"2\",\n" + "dst_indicator: \"E\",\n" + "city: \"Paris\",\n" + "country: \"France\",\n" + "country_code: \"FR\",\n" + "region: \"TC1\",\n" + "listing_display: \"true\",\n" + "pseudonyms: \"\"\n" + "},\n" + "ORY: {\n" + "id: \"33539\",\n" + "airport_name: \"Orly\",\n" + "latitude: \"48.7167\",\n" + "longitude: \"2.3833\",\n" + "timezone: \"2\",\n" + "dst_indicator: \"E\",\n" + "city: \"Paris\",\n" + "country: \"France\",\n" + "country_code: \"FR\",\n" + "region: \"TC1\",\n" + "listing_display: \"true\",\n" + "pseudonyms: \"\"\n" + "},\n" + "LBG: {\n" + "id: \"123425\",\n" + "airport_name: \"Le Bourget\",\n" + "latitude: \"48.969444\",\n" + "longitude: \"2.441389\",\n" + "timezone: \"1\",\n" + "dst_indicator: \"E\",\n" + "city: \"Paris\",\n" + "country: \"France\",\n" + "country_code: \"FR\",\n" + "region: \"TC1\",\n" + "listing_display: \"true\",\n" + "pseudonyms: \"\"\n" + "}}"; 

使用gson，您可以检索密钥名称，如下所示：

  Gson gson = new Gson(); Object o = gson.fromJson(str, Object.class); List keys = new ArrayList(); Collection values = null; if (o instanceof Map) { Map map = (Map) o; keys.addAll(map.keySet()); // collect keys at current level in hierarchy values = map.values(); } else if (o instanceof Collection) { values = (Collection) o; } System.out.println(keys);// [CDG, ORY, LBG] for (int i = 0; i < keys.size(); i++) { System.out.println(keys.get(i)); } 

并使用java-json，您可以执行以下操作：

  JSONObject jsonObject = new JSONObject(str); String[] names = jsonObject.getNames(jsonObject); for (int i = 0; i < names.length; i++) { System.out.println(names[i]);// all names are printed here : LBG,ORY, etc // then parsing the names accordingly.. JSONObject jsonObject1 = jsonObject.getJSONObject(names[i]); System.out.println(jsonObject1.getString("city")); } 

从url获取json：

  public static String connectionGet(String url, String parameter) throws MalformedURLException, ProtocolException, IOException { URL url1 = new URL(url); HttpURLConnection request1 = (HttpURLConnection) url1.openConnection(); request1.setRequestMethod("GET"); request1.connect(); String responseBody = convertStreamToString(request1.getInputStream()); return responseBody; } String str = connectionGet("http://www.cleartrip.com/common/json/airports.json", ""); 

Answer 2

 ArrayList keylist = new ArrayList(); for ( String key : your_map_name.keySet() ) { System.out.println( key ); keylist.add(""+key); } 

所以从这里你可以获得没有任何tnsn的地图的所有关键，只需将它添加到arraylist所以只需传递键作为arraylist索引值并获得map的值所以你不用担心json键无论它是什么存储在arraylist索引中。 🙂