1

这是Asynctask方法：

public class Read extends AsyncTask<String, Integer, String> {
    ProgressDialog dialog;

    @Override
    protected void onPreExecute() {
        // TODO Auto-generated method stub
    }

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

}

以及相关的 lastTweet 方法：

public JSONObject lastTweet(String username)
        throws ClientProtocolException, IOException, JSONException {
    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 timeline = new JSONArray(data);
        JSONObject last = timeline.getJSONObject(0);
        return last;
    }
}

所有这些代码都工作正常。目前没有任何问题。但是，我想做一些小的修补。当HTTP 传输期间连接丢失时，RemoteException 会抛出 a 并且应用程序崩溃。

我试图在Async方法中处理异常但无法这样做。

有什么办法可以处理这种异常？

4

1 回答 1

1

您可以在执行之前检查网络连接Read AsyncTask

1.检查连接

if(ifConnectionIsAvailable)
    new Read().execute();

2. 设置连接超时

HttpGet get = new HttpGet(url.toString());
HttpParams httpParameters = new BasicHttpParams();
int timeoutConnection = 3000;// in milliseconds 
HttpConnectionParams.setConnectionTimeout(httpParameters, timeoutConnection);
int timeoutSocket = 5000;
HttpConnectionParams.setSoTimeout(httpParameters, timeoutSocket);
DefaultHttpClient httpClient = new DefaultHttpClient(httpParameters);
HttpResponse r = httpClient.execute(get);
于 2013-07-14T19:54:15.650 回答