这是回复 我想知道我该如何处理这样的JSON响应? 这是一个JSONArray,但没有一个名字 这里是回应:JSON:如何处理这样的JSON响应?
[[84,"sinat","[email protected]"],[88,"msn","[email protected]"],[89,"163t","[email protected]"],[90,"mail","[email protected]"],[93,"mail","[email protected]"]]
这是回复 我想知道我该如何处理这样的JSON响应? 这是一个JSONArray,但没有一个名字 这里是回应:JSON:如何处理这样的JSON响应?
[[84,"sinat","[email protected]"],[88,"msn","[email protected]"],[89,"163t","[email protected]"],[90,"mail","[email protected]"],[93,"mail","[email protected]"]]
数据不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,这是一个JSONArray,我错过了,谢谢你的提议。我正在尝试你的方法:) –
你好,我不知道如何获得像“sinat”这样的详细信息?它还是一个JSONArray? –
不,你可以得到像'json_array.getJSONArray(i).getString(1)' –
'JSONArray ARR =新JSONArray( “[[..]]”); (int i; i
Selvin
是啊,非常感谢你:) –