4レベルのネストしたJSON配列で名前を抽出できません

Question

レストランの詳細を表示するアンドロイドアプリを作成しようとしています。私は引数として文字列jsonを取るparseJSON関数を持っています。今私がしようとしていることは、下のjsonにある個々のレストランの名前を示すことです。私は個々のレストラン（3レベルの深さ）を抽出してlog.dコンソールから出力することができますが、個々のレストランの名前を抽出することはできません。コードはこちら

public void parseJSON(String jsonQuery){ 
     //Parsing JSON 
     try { 
      JSONObject jsonObject = new JSONObject(jsonQuery); 
      if (jsonObject != null){ 
       JSONArray restaurants = jsonObject.getJSONArray("restaurants"); 
       for (int i = 0; i < restaurants.length(); i++){ 
        JSONObject restaurant = restaurants.getJSONObject(i); 
        if (restaurant != null){ 
         String tempName = restaurant.getString("name"); 
         Log.d(TAG, "Restaurant name: " + tempName); 
        } 
       } 
      } 

     } catch (JSONException e) { 
      Log.d(TAG, "Exception catched: " + e); 
      e.printStackTrace(); 
     } 
    } 

ここは私のjsonファイルです。私は取得しています

{ 
    "results_found": 1281966, 
    "results_start": 0, 
    "results_shown": 2, 
    "restaurants": [ 
    { 
     "restaurant": { 
     "R": { 
      "res_id": 16607974 
     }, 
     "apikey": "123456789", 
     "id": "16607974", 
     "name": "Bassine Specialty Cheese", 
     "url": "https://www.zomato.com/bass-vic/bassine-specialty-cheese-bass?utm_source=api_basic_user&utm_medium=api&utm_campaign=v2.1", 
     "location": { 
      "address": "2125 Bass Hwy", 
      "locality": "Bass", 
      "city": "Bass", 
      "city_id": 1341, 
      "latitude": "-38.4833750000", 
      "longitude": "145.4670320000", 
      "zipcode": "3995", 
      "country_id": 14, 
      "locality_verbose": "Bass, Bass" 
     }, 
     "switch_to_order_menu": 0, 
     "cuisines": "Ice Cream", 
     "average_cost_for_two": 7, 
     "price_range": 1, 
     "currency": "$", 
     "offers": [], 
     "thumb": "", 
     "user_rating": { 
      "aggregate_rating": "2.9", 
      "rating_text": "Average", 
      "rating_color": "FFBA00", 
      "votes": "5" 
     }, 
     "photos_url": "https://www.zomato.com/bass-vic/bassine-specialty-cheese-bass/photos?utm_source=api_basic_user&utm_medium=api&utm_campaign=v2.1#tabtop", 
     "menu_url": "https://www.zomato.com/bass-vic/bassine-specialty-cheese-bass/menu?utm_source=api_basic_user&utm_medium=api&utm_campaign=v2.1&openSwipeBox=menu&showMinimal=1#tabtop", 
     "featured_image": "", 
     "has_online_delivery": 0, 
     "is_delivering_now": 0, 
     "deeplink": "zomato://restaurant/16607974", 
     "has_table_booking": 0, 
     "events_url": "https://www.zomato.com/bass-vic/bassine-specialty-cheese-bass/events#tabtop?utm_source=api_basic_user&utm_medium=api&utm_campaign=v2.1", 
     "establishment_types": [] 
     } 
    }, 
    { 
     "restaurant": { 
     "R": { 
      "res_id": 17649424 
     }, 
     "apikey": "3d93604b4a84d85f374f39ea3b644132", 
     "id": "17649424", 
     "name": "Schobels' Restaurant", 
     "url": "https://www.zomato.com/columbus-tx/schobels-restaurant-columbus?utm_source=api_basic_user&utm_medium=api&utm_campaign=v2.1", 
     "location": { 
      "address": "2020 Milam St 78934", 
      "locality": "Columbus", 
      "city": "Columbus", 
      "city_id": 9241, 
      "latitude": "29.6965000000", 
      "longitude": "-96.5405000000", 
      "zipcode": "78934", 
      "country_id": 216, 
      "locality_verbose": "Columbus, Columbus" 
     }, 
     "switch_to_order_menu": 0, 
     "cuisines": "German, Southern", 
     "average_cost_for_two": 25, 
     "price_range": 2, 
     "currency": "$", 
     "offers": [], 
     "thumb": "", 
     "user_rating": { 
      "aggregate_rating": "4.0", 
      "rating_text": "Very Good", 
      "rating_color": "5BA829", 
      "votes": "164" 
     }, 
     "photos_url": "https://www.zomato.com/columbus-tx/schobels-restaurant-columbus/photos?utm_source=api_basic_user&utm_medium=api&utm_campaign=v2.1#tabtop", 
     "menu_url": "https://www.zomato.com/columbus-tx/schobels-restaurant-columbus/menu?utm_source=api_basic_user&utm_medium=api&utm_campaign=v2.1&openSwipeBox=menu&showMinimal=1#tabtop", 
     "featured_image": "", 
     "has_online_delivery": 0, 
     "is_delivering_now": 0, 
     "deeplink": "zomato://restaurant/17649424", 
     "has_table_booking": 0, 
     "events_url": "https://www.zomato.com/columbus-tx/schobels-restaurant-columbus/events#tabtop?utm_source=api_basic_user&utm_medium=api&utm_campaign=v2.1", 
     "establishment_types": [] 
     } 
    } 
    ] 
} 

エラー： org.json.JSONException：名前なし値

Answer 1

あなたjsonQueryがrestaurantという名前JsonObject内JsonObjectが含まれているので、あなたがJSONObject restaurantsJSONObject =restaurant.getJSONObject("restaurant");同様JSONObject restaurant = restaurants.getJSONObject(i);

からJsonObjectを取得する必要があります

オブジェクトから名前を取得する例を以下に試してください

try { 
    JSONObject jsonObject = new JSONObject(jsonQuery); 
    if (jsonObject != null){ 
     JSONArray restaurants = jsonObject.getJSONArray("restaurants"); 
     for (int i = 0; i < restaurants.length(); i++){ 
      JSONObject restaurant = restaurants.getJSONObject(i); 
      JSONObject restaurantsJSONObject =restaurant.getJSONObject("restaurant"); 
      if(restaurantsJSONObject.has("name")) { 
       String tempName = restaurantsJSONObject.getString("name"); 
       Log.d(TAG, "Restaurant name: " + tempName); 
      } 
     } 
    } 
} catch (JSONException e) { 
    Log.d(TAG, "Exception catched: " + e); 
    e.printStackTrace(); 
}