接收来自PHP文件的HTTP POST回显响应（发送POST效果很好，我无法理解接收）

Question

因此，正如标题所示，我的问题是获得对我正在制作的HTTP POST的响应。应该发生的是，我发送了一堆变量，PHP检查数据库中的变量，并将结果发回给我（作为页面的回声）。

以下是android代码：

 public class CheckChallenge extends AsyncTask<String, Void, String> 
  {
    @Override
    protected String doInBackground(String... urls) 
    {
      String response = "";
      try
      {
        URL = urls[0];
   ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(4);
   nameValuePairs.add(new BasicNameValuePair("victim",NetPlay.myId)); 
   // need to return these to an array
   nameValuePairs.add(new BasicNameValuePair("rival",rivalid));


        nameValuePairs.add(new BasicNameValuePair("word","null"));
        nameValuePairs.add(new BasicNameValuePair("won","0"));

          HttpClient httpclient = new DefaultHttpClient();
          HttpPost httppost = new      
          HttpPost("http://www.hanged.comli.com/check-rival.php");
          httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

          HttpResponse execute = httpclient.execute(httppost);
          HttpEntity entity = execute.getEntity();

          //InputStream is = entity.getContent();

          //mText.setText(is.toString());

         Log.i("postData", execute.getStatusLine().toString());
         //HttpEntity entity = response.getEntity();

      }
      catch(Exception e)
      {
              Log.e("log_tag", "Error in http connection"+e.toString());
      }
            return response;
        } 

    @Override
    protected void onPostExecute(String result) 
    {
        // CHECK ENTIRE DATABASE FOR MY ID //
        // IF MY ID IS THERE THEN THAT MEANS IVE CHALLENGED SOMEONE //


    }

  }

这里是PHP，我认为它是可以的，只是为了完整性而包括它：$connect=mysql\u connect（$mysql\u host“，$mysql\u user“，$mysql\u password”）或die（$mysql\u password”）；mysql_选择_数据库（$mysql_数据库“，$connect）或die（“无法选择数据库”）；会话_start（）；

$victim = $_POST['victim'];
$rival = $_POST['rival'];
$word = $_POST['word'];
$won = $_POST['won'];

$query = "SELECT rival FROM currentgames";
$result = mysql_query($query);

 if (!$result) 
 {
echo "Could not successfully run query ($sql) from DB: " . mysql_error();
exit;
}

if (mysql_num_rows($result) == 0) 
{
echo "No rows found, nothing to print so am exiting";
exit;
}

 while ($row = mysql_fetch_assoc($result)) 
{
 echo $row["rival"];
}

如果能帮上忙，我将不胜感激，尽量让我了解所有这些HTTP帖子的内容。

Answer 1

发送HTTP请求并读回HTTP响应的示例：

String res = "";
String url = "http://www.domain.com/youscript.php";
URL urlObj = new URL(url);
URLConnection lu = urlObj.openConnection();


// Send data - if you don't need to send data 
// ignore this section and just move on to the next one
String data = URLEncoder.encode("yourdata", "UTF-8");
lu.setDoOutput(true);
OutputStreamWriter wr = new OutputStreamWriter(lu.getOutputStream());
wr.write(data);
wr.flush();

// Get the response
BufferedReader rd = new BufferedReader(new InputStreamReader(lu.getInputStream()));
String line = "", res = "";
while ((line = rd.readLine()) != null) {
  res += line;
}

wr.flush();
wr.close();
System.out.println(res);