這是回覆 我想知道我該如何處理這樣的JSON響應? 這是一個JSONArray,但沒有一個名字 這裏是迴應:JSON:如何處理這樣的JSON響應?
[[84,"sinat","[email protected]"],[88,"msn","[email protected]"],[89,"163t","[email protected]"],[90,"mail","[email protected]"],[93,"mail","[email protected]"]]
這是回覆 我想知道我該如何處理這樣的JSON響應? 這是一個JSONArray,但沒有一個名字 這裏是迴應:JSON:如何處理這樣的JSON響應?
[[84,"sinat","[email protected]"],[88,"msn","[email protected]"],[89,"163t","[email protected]"],[90,"mail","[email protected]"],[93,"mail","[email protected]"]]
數據不JSONObject
但JSONArray
!
String json_value = '[[84,"sinat","[email protected]"],[88,"msn","[email protected]"],[89,"163t","[email protected]"],[90,"mail","[email protected]"],[93,"mail","[email protected]"]]';
JSONArray json_array = new JSONArray(json_value);
穿行它像
for(int i = 0; i < json_array.length(); i++){
Log.d("Current index: "+i,"Current value: "+json_array.getJSONArray(i).toString());
}
是的,這是一個JSONArray,我錯過了,謝謝你的提議。我正在嘗試你的方法:) –
你好,我不知道如何獲得像「sinat」這樣的詳細信息?它還是一個JSONArray? –
不,你可以得到像'json_array.getJSONArray(i).getString(1)' –
'JSONArray ARR =新JSONArray( 「[[..]]」); (int i; i
Selvin
是啊,非常感謝你:) –