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我想消费一个休息服务，下载json并将其放入一个对象中，然后返回它，但该对象总是返回我null，这是类：

public class JSONParser {

static InputStream is = null;
static JSONObject jObj = null;
static String json = "";
Context ctx;

// constructor
public JSONParser(Context ctx) {

    this.ctx = ctx;

}

public JSONObject getJSONFromUrl(String url) {

    AsyncjSONTask task = new AsyncjSONTask();

    task.execute(url);

    return jObj;

}

private class AsyncjSONTask extends AsyncTask<String, Void, JSONObject>{



    @Override
    protected JSONObject doInBackground(String... params) {

        String url = params[0];
        InputStream is = null;
        // Making HTTP request
                try {
                    // defaultHttpClient
                    DefaultHttpClient httpClient = new DefaultHttpClient();
                    HttpPost httpPost = new HttpPost(url);

                    HttpResponse httpResponse = httpClient.execute(httpPost);
                    HttpEntity httpEntity = httpResponse.getEntity();
                    is = httpEntity.getContent();           

                } catch (UnsupportedEncodingException e) {
                    e.printStackTrace();
                } catch (ClientProtocolException e) {
                    e.printStackTrace();
                } catch (IOException e) {
                    e.printStackTrace();
                }
                JSONObject jObjOut = null;
                try {
                    BufferedReader reader = new BufferedReader(new InputStreamReader(is, "iso-8859-1"), 8);
                    StringBuilder sb = new StringBuilder();
                    String line = null;
                    while ((line = reader.readLine()) != null) {
                        sb.append(line + "\n");
                    }
                    is.close();
                    json = sb.toString();
                } catch (Exception e) {
                    Log.e("Buffer Error", "Error converting result " + e.toString());
                }

                // try parse the string to a JSON object
                try {
                    jObjOut = new JSONObject(json);

                } catch (JSONException e) {
                    Log.e("JSON Parser", "Error parsing data " + e.toString());
                }
        return jObjOut;
    }

    @Override
    protected void onPostExecute(JSONObject jObjIn) {

        jObj = jObjIn;
    }

}

}

如果还有其他消费休息服务的方式，请告诉我。

4

1 回答 1

1
  1. 请确保您要进行 HTTP POST 而不是 GET。
  2. 在阅读响应之前，最好检查一下 HTTP 响应状态是什么
  3. 不要将您的异步代码包装在非异步类中？似乎您感到困惑并把它称为好像它不是异步的。

您的 JSONParser 类假设 AsyncTask 不是异步的，而实际上它是。这是一个示例，说明您将如何做您正在尝试做的事情：

public class MainActivity extends Activity {

    @Override
    public void onCreate(Bundle savedInstanceState) {
        new AsyncJsonTask(this).execute();

    }

    public void doSomethingWithTheResult(JsonObject result) {
        // Show the result on the View or do whatever with it.
    }

    private class AsyncJsonTask extends AsyncTask<String, Void, JsonObject> {

        private MainActivity _activity;

        public AsyncJsonTask(MainActivity activity) {
            this._activity = activity;
        }

        @Override
        protected JsonObject doInBackground(String... params) {
            JsonObject outputObject = null;

            // Call your web service to return the output 
            // ...

            return outputObject
        }

        @Override
        protected void onPostExecute(JsonObject result) {
            _activity.doSomethingWithTheResult(result);
        }

    }
}
于 2012-12-04T23:32:17.943 回答