Question

我的Android应用程序中有这个JSON字符串

[{"tel1":"999","tel2":"0790000000","tel3":"","interval":"60","deleteLocal":"1","id":"2"}]

如何将其解析为JsonArray然后获取值，例如tel1？

Answer 1

try
    {
       JSONArray jArray= new JSONArray(output);
       JSONObject jsonObject=jsonArray.getJSONObject(0);
       String tel= jsonObject.getString("tel1");
    }catch(Exception e)
    {
     //error parsing response.
    }

Answer 2

喜欢这个

{ - 代表JsonObject

[ - 代表JsonArray

  try
        {
        JSONArray jArray= new JSONArray(output);
               JSONObject menuObject = JSONArray.getJSONObject(0);  
               String tel= menuObject.getString("tel1");
        }catch(Exception e)
        {

Log.e("d",e.getMessage());
        }

Answer 3

试试这个

try {
        JSONArray jsonArray = new JSONArray(responsestring);
        for (int i = 0; i < jsonArray.length(); i++) {

            JSONObject jobj = jsonArray.getJSONObject(i);
            String strtel1 = jobj.getString("tel1");

            Log.i("value tel1 : ", strtel1 + "");
        }
    } catch (JSONException e) {
        // TODO: handle exception
    }

Answer 4

试试这个。

JSONArray arr = new JSONArray(result);
JSONObject jObj = arr.getJSONObject(0);
String tel1 = jObj.getString("tell");

Answer 5

试试这个：

try {
    JSONArray jsonArray= new JSONArray(yourString);
    if(jsonArray != null && jsonArray.length() > 0) {
        for(int i=0; i< jsonArray.length(); i++) {
            JSONObject json = jsonArray.getJSONObject(i);
            if(json != null) {
                String tel1 = json.optString("tel1", "default_value");
                Log.d("PARSE", "tel1  : "+tel1);
            }
        }
    }
}catch(Exception e)
{
}

android中的JSON String如何获取值

5 个答案: