這是我從一個URL獲取JSONJSONarray分析顯示沒有價值發現
{
responce: "success",
data: [
{
Breakfast: [
{
food_interval: "Breakfast",
id: "1",
menu_creator_id: "1",
item_code: "13",
food_interval_category_id: "1",
food_interval_category: "Main Item",
item_name: "Methi thepal",
item_quantity: "1",
unit_id: "1",
unit_desc: "Number"
},
{
food_interval: "Breakfast",
id: "2",
menu_creator_id: "1",
item_code: "14",
food_interval_category_id: "2",
food_interval_category: "Cereals",
item_name: "Museli",
item_quantity: "1",
unit_id: "7",
unit_desc: "Bowl "
},
{
food_interval: "Breakfast",
id: "3",
menu_creator_id: "1",
item_code: "15",
food_interval_category_id: "8",
food_interval_category: "Bread",
item_name: "Brown Bread",
item_quantity: "2",
unit_id: "1",
unit_desc: "Number"
},
{
food_interval: "Breakfast",
id: "4",
menu_creator_id: "1",
item_code: "16",
food_interval_category_id: "21",
food_interval_category: "Butter Cheese",
item_name: "Cheddar Cheese",
item_quantity: "1",
unit_id: "3",
unit_desc: "Gram"
}
]
}
]
}
代碼,我曾嘗試:
String result = response.body().string();
JSONObject jsonObject = new JSONObject(result);
JSONArray foodintervalarray = jsonObject.getJSONArray("data");
for(int i = 0 ; i < foodintervalarray.length(); i++){
JSONObject jsonObject1 = foodintervalarray.getJSONObject(i);
JSONArray breakfast = jsonObject1.getJSONArray("Breakfast");
for(int j = 0 ; j < breakfast.length(); j++){
JSONObject jsonObject2 = breakfast.getJSONObject(j);
String breakFastMenu = jsonObject2.getString("food_interval");
Log.i("breakFastMenu","breakFastMenu "+breakFastMenu);
}
但我正在逐漸:未找到早餐數組值。 數據數組後有一個{,所以我需要調用JsonArray早餐前調用JsonObject? 有人建議我做什麼
感謝
http://dieto.vm1.in/api/menu_display_2d.php?date=2016/08/22&diet_type=6&food_category=1&class_type=1 當我jsonlint測試它,它讓我看到正確的,它也說有效,但在我的瀏覽器中顯示不同,爲什麼這樣? –
這是有點正確的json.w8。我會查的。 – Vyacheslav
@narahari_arjun,已更新。檢查你如何檢索你的json數據。這是對的嗎。 – Vyacheslav