如何在android中发送发布数据并检索响应正文？答案

【问题标题】：How to send post data and retrieve the response body in android?如何在android中发送发布数据并检索响应正文？
【发布时间】：2014-03-11 02:07:13
【问题描述】：

代码可以找到，但它没有返回正确的 HTTPRESPONSE，它假设返回成功！作为回应，而是返回一些乱码。这是我的代码，请帮助我，我尝试了很多东西，但没有任何反应，请指教。

       public class PingServerActivity extends Activity {

   private Button btn6;
    //private EditText value;
     protected String name;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_ping_server);

    btn6 = (Button) findViewById(R.id.btn_6);
    //value = (EditText) findViewById(R.id.textV5);

    pingButton();

}

public void pingButton() {

    btn6.setOnClickListener(new OnClickListener() {

        @Override
        public void onClick(View v) {
            // this button would send the request to the server
            new PingPostTask().execute(new String[] {name}); 

        }
    });

}

private class PingPostTask extends AsyncTask<String, Void, String>{

     @Override
        protected String doInBackground(String... params)   
        {           
            BufferedReader inBuffer = null;
            String url = "http://ec2-54-243-205-92.compute-1.amazonaws.com/Tests/ping.php";
            String result = "fail";
            try {
                HttpClient httpClient = new DefaultHttpClient();
                HttpPost request = new HttpPost(url);
                List<NameValuePair> postParameters = new ArrayList<NameValuePair>();
                postParameters.add(new BasicNameValuePair("Password", "EGOT"));

                UrlEncodedFormEntity formEntity = new UrlEncodedFormEntity(
                        postParameters);

                request.setEntity(formEntity);
                HttpResponse response = httpClient.execute(request);//btw, i'm NOT very sure if this will work LOL
                       result = response.toString();

            } catch(Exception e) {
                // Do something about exceptions
                result = e.getMessage();
            } finally {
                if (inBuffer != null) {
                    try {
                        inBuffer.close();
                    } catch (IOException e) {
                        e.printStackTrace();
                    }
                }
            }
            return  result;
        }

        protected void onPostExecute(String result)
        {       
            //textView.setText(page); 
            Toast toast = Toast.makeText(getApplicationContext(), result, Toast.LENGTH_LONG);
          toast.show();
        }


}

@Override
public boolean onCreateOptionsMenu(Menu menu) {

    // Inflate the menu; this adds items to the action bar if it is present.
    getMenuInflater().inflate(R.menu.ping_server, menu);
    return true;
}

}

【问题讨论】：

标签： android asynchronous android-asynctask http-post httpresponse

【解决方案1】：

你不能简单地将 HTTPResponse 转换为字符串，你需要将它读入输入流或类似的，给

替换：

result = response.toString();

与

result = EntityUtils.toString(response);

去吧。

【讨论】：

嘿@xBroak 你能解释一下我不明白的地方

【解决方案2】：

响应有时不仅是String类型，也可以是二进制，建议你从inputstream中检索内容

HttpEntitiy entitiy = response.getEntity();
Inputstream inputstream = entitiy.getContent();
//do what you want, e.g. convert to string or decode as bitmap

【讨论】：