2

我得到了这个 Logcat：

08-14 11:42:55.923: W/System.err(1137): org.json.JSONException: Value {"error":0,"success":0,"tag":"helloworld"} of type org.json.JSONObject cannot be converted to JSONArray

我该如何解决这个问题？

08-14 11:42:55.923: W/System.err(1137):     at org.json.JSON.typeMismatch(JSON.java:111)

我使用的代码是：

public class MainActivity extends Activity {

    TextView tv;
    HttpClient client;
    JSONObject json;
final static String URL = "http://192.168.1.9/sumit/hello.php";
    //final static String URL = "http://json.org/example.html";
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

        tv = (TextView) findViewById(R.id.tv1);
        client = new DefaultHttpClient();

        Log.e("my","check");
        new Read().execute("tag");
//      StrictMode.ThreadPolicy policy = new StrictMode.ThreadPolicy.Builder()
//              .detectAll().penaltyLog().build();
//      StrictMode.setThreadPolicy(policy);

    }

    public JSONObject myData() throws ClientProtocolException, IOException,
            JSONException {

        StringBuilder url = new StringBuilder(URL);
        HttpGet get = new HttpGet(url.toString());
        HttpResponse r = client.execute(get);
        int status = r.getStatusLine().getStatusCode();
        if (status == 200) {
            HttpEntity e = r.getEntity();
            String data = EntityUtils.toString(e);
            JSONArray datastream = new JSONArray(data);
            JSONObject message = datastream.getJSONObject(0);
            return message;
        } else {
            Toast.makeText(MainActivity.this, "error encountered",
                    Toast.LENGTH_SHORT).show();
            return null;
        }

    }

    class Read extends AsyncTask<String, Integer, String> {

        @Override
        protected String doInBackground(String... params) {
            // TODO Auto-generated method stub
            try {
                json = myData();

                                    return json.getString(params[0]);



            } catch (ClientProtocolException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            } catch (IOException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            } catch (JSONException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }

            return null;
        }

        @Override
        protected void onPostExecute(String result) {
            // TODO Auto-generated method stub
            tv.setText(result);

        }

    }

}

生成的 JSON 代码：

{"tag":"hello","success":0,"error":0}

用于生成的代码：

<?php


    // response Array
    $response = array("tag" => helloworld, "success" => 0, "error" => 0);


   echo json_encode($response);


?>
4

1 回答 1

2

因为你得到的不是JSONArray——而是JSONObject

JSONArray 在您的情况下将如下所示：

[{"error":0,"success":0,"tag":"helloworld"}]

它是一个包含一个元素的数组（即JSONObject，它包含字段）。

我不太了解php（而且我不知道它是如何json_encode工作的），但试试这个：

<?
    $response = array(array("tag" => helloworld, "success" => 0, "error" => 0));
    echo json_encode($response);
?>
于 2012-08-14T09:05:07.980 回答