Question

我正在尝试解析URL以获取JSON响应和该响应中的特定值。我没有示例代码。请给我一个简单的解决方案。下面我发布了我的网址和回复。我想获得“学校”，“名称”和结果值。

http://sample.com/login/username/ <username> /password <password>?  

{
    "response":{
                "School":"SBOA",
                "Name":"Anitha",
                "Class":"Tenth",
              },
                "Result":"Good",
}

我的代码：

public class MainActivity extends Activity {
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

        TextView output = (TextView) findViewById(R.id.textView1);
        String strJson="URL";
        String data = "";
        System.out.println(strJson);

        try {
            JSONObject  jsonRootObject = new JSONObject(strJson);

            //Get the instance of JSONArray that contains JSONObjects
            JSONArray jsonArray = jsonRootObject.optJSONArray("response");
            System.out.println(jsonRootObject);

            //Iterate the jsonArray and print the info of JSONObjects
            for(int i=0; i < jsonArray.length(); i++){
                JSONObject jsonObject = jsonArray.getJSONObject(i);

            //    int id = Integer.parseInt(jsonObject.optString("id").toString());
                String name = jsonObject.optString("School").toString();
            //   float salary = Float.parseFloat(jsonObject.optString("salary").toString());

             //   data += "Node"+i+" : \n id= "+ id +" \n Name= "+ name +" \n Salary= "+ salary +" \n ";*/
            }
            //output.setText(data);
        } catch (JSONException e) {e.printStackTrace();}
    }
}

Answer 1

如果你有一个名为json的JSONObject，那么按照这个来获取学校价值

try{
jsonRootObject .getJSONObject("response").getString("School");
}catch(JSONException e)
{
e.printStackTrace();
}

Answer 2

要准确读取JSON，请使用：

/** Verify that your strJson string contains this:
 * {
 *   "response":{
 *               "School":"SBOA",
 *               "Name":"Anitha",
 *               "Class":"Tenth",
 *             },
 *               "Result":"Good",
 * }
 */
String strJson = ??;
Log.d("TAG", "strJson: " + strJson);
try {
    JSONObject jsonRootObject = new JSONObject(strJson);
    JSONObject response = jsonRootObject.getJsonObject("response");
    String schoolString = response.getString("School");
    String nameString = response.getString("Name");
    String classString = response.getString("Class");

    String result = jsonRootObject.getString("Result");
} catch(JSONException e) {
    Log.e("TAG", "Error reading json: " + jsonRootObject.toString());
}

如何使用Android解析嵌套的JSON？

2 个答案: