如何在java上获取get请求的响应

Question

我正在做一个Android应用程序，我只是做一个获取http请求，我只是想知道我做得不好。

这是我的要求：

try {
    String encodedUserNAme = idDispositivo;
    encodedUserNAme =  URLEncoder.encode(encodedUserNAme, "utf-8");
    String url = Constants.URL_OBTENER_GRUPO + "?" + "user="+ idDispositivo;
    Log.w("ObtenerIDGrupo "," url: "+url);



    HttpClient client = Constants.getHttpClient();
    HttpGet httpget;
    try {
        httpget = new HttpGet(new URI(url));

        try{
            HttpResponse response = client.execute(httpget);
            String responseBody = EntityUtils.toString(response.getEntity());

                    Log.w("ObtenerIDGrupo "," RESPONSEBODY= "+responseBody);
                    idGrupo = responseBody;
                    Log.w("ObtenerIDGrupo "," idGrupo= "+responseBody);

        }catch(ClientProtocolException e){
            e.printStackTrace();
        }catch(IOException e){
            e.printStackTrace();
        }
    } catch (URISyntaxException e1) {
        e1.printStackTrace();
    }
} catch (IOException e) {
    Log.w("IOException" + e);
    e.printStackTrace();
}

这是我的回答：

我不知道为什么回答是这样的。 请有人帮忙吗？

由于

Answer 1

您无法像这样操纵响应，请尝试以下代码：

InputStream is = response.getEntity().getContent();        
BufferedReader bufferedReader = new BufferedReader(
            new InputStreamReader(is));

StringBuilder str = new StringBuilder();
String line = null;

while ((line = bufferedReader.readLine()) != null) {
    str.append(line + "\n");
}
responseBody = str.toString();
Log.w("ObtenerIDGrupo "," RESPONSEBODY= "+responseBody);

Answer 2

试试这个

  URL url = new URL(yoururl);
         InputStream input=url.openStream();
         BufferedInputStream bis=new BufferedInputStream(input);
         ByteArrayBuffer baf=new ByteArrayBuffer(1000);
         while((k=bis.read())!=-1)
         {
         baf.append((byte)k);
         }
        String responsedata=new String(baf.toByteArray());