2011-09-30 47 views

回答

1

數據不JSONObjectJSONArray

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()); 
} 
+0

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:) –

+0

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? –

+0

不,你可以得到像'json_array.getJSONArray(i).getString(1)' –