Question

背景：我是android编程的新手。我想简单地向本地服务器发送一个http get请求。

我想将此请求作为参数传递给该请求，并希望在json中获得返回。这个问题我无法在主线程上执行它。我怎么能这样做？

以下是我的尝试：

主要课程：

itemsAdapter.add(get.getName(device.getName()));

在同一档案中单独分类：

private class httpGet extends AsyncTask<Editable, Void, Integer> {
        protected String doInBackground(Editable... params) {
            Editable editable = params[0];
            return getName(editable.toString());
        }
        final String getName(String btName) {
            HttpResponse response = null;
            String result = "";
            try {
                HttpClient client = new DefaultHttpClient();
                HttpGet request = new HttpGet();
                URI website = new URI("http://192.168.1.105/getName.php?q=" + btName);
                request.setURI(website);
                response = client.execute(request);
                // Convert String to json object
                JSONObject json = new JSONObject(response.toString());

                // get LL json object
                JSONObject json_Name = json.getJSONObject("Name");

                // get value from LL Json Object
                name = json_Name.getString("value"); //<< get value here
            } catch (URISyntaxException e) {
                e.printStackTrace();
            } catch (ClientProtocolException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            } catch (IOException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            } catch (JSONException e) {
                e.printStackTrace();
                // Do something to recover ... or kill the app.
            }

            return result;
        }

        protected void onPostExecute(Integer result) {
            // here you have the result
        }

我不确定这是否是完成此任务的好方法。我也不知道我怎么称呼它。

Answer 1

您应该了解asyncTask的工作原理。在DoInBackground中，您应该将代码引用到HTTPRequest。我建议使用方法来提高对代码的理解。以下是我的某个应用的示例：

 public String query(String uri) {
    HttpClient cliente = new DefaultHttpClient();
    HttpContext contexto = new BasicHttpContext();
    HttpPost httpPost = new HttpPost(uri);
    HttpResponse response = null;
    String resultado=null;

    try {
        List<NameValuePair> params = new ArrayList<NameValuePair>(2);
        params.add(new BasicNameValuePair("dato", cod_restaurante));
        httpPost.setEntity(new UrlEncodedFormEntity(params));
        response = cliente.execute(httpPost, contexto);
        HttpEntity entity = response.getEntity();
        resultado = EntityUtils.toString(entity, "UTF-8");


    } catch (Exception e) {
        // TODO: handle exception
    }
    return resultado;
}

 private class MyAsyncTask extends AsyncTask<String, Integer, String>{

    @Override
    protected String doInBackground(String... params) 
    {
        result=query(params[0]);
        return result;
    }

    protected void onPostExecute(final String resultadoDoInBackground)
    { 
      //here put the code to modify the UI
     }
 }

然后在您的活动onCreate（）方法中执行Asynktask。

 new MyAsyncTask().execute("    ");

您可以在此处阅读有关AsyncTask的更多信息： AsyncTask Android Developer

Answer 2

AsyncTask允许您在不同的线程中执行后台操作，而无需操作线程/处理程序。

应该是这样的：

private class httpGet extends AsyncTask<ParamForDoInBackground, ParamForOnProgressUpdate, ParamForOnPostExecute> {
     protected Long doInBackground(ParamForDoInBackground... urls) {
        // do the request here
     }

     protected void onProgressUpdate(ParamForOnProgressUpdate progress) {
        // if you need to show any progress of the 
        // request from doInBackground
     }

     protected void onPostExecute(ParamForOnPostExecute result) {
        // this method will run when doInBackground
        // is done executing
     }
}

然后你可以执行AsyncTask：

new httpGet().execute(ParamForDoInBackground);

您可以将以下内容用作参考：AndroidBackgroundProcessing和Android Developer AsyncTask

Android - 单独线程上的HTTP GET

2 个答案: