recieving HTTP从PHP文件发布回送应答（发送帖子做工精细，它的接收，我想不通） [英] recieving HTTP POST echo response from a PHP file (sending the POSTS works fine, it's the receive that I can't figure out)

问题描述

因此​​，作为标题暗示我的问题越来越为我做一个HTTP POST的响应。
应该怎样发生的事情是我送了一堆的变数，PHP的检查数据库他们和发回给我的结果（如回声页面）。

So as the title suggest my problem is getting a response to a HTTP POST I'm making. What SHOULD be happening is I send a bunch of variables, the PHP checks the database for them and sends back to me the result (as an echo to the page).

下面是Android code：

Here is the android code:

 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 //


    }

  }

下面是我认为这是确定只包括本的完整性PHP：
     $连接= mysql_connect（$ MYSQL_HOST，$ mysql_user，$ mysql_password）或死亡（无法连接）;
    mysql_select_db（$ mysql_database，$连接）或死亡（不能选择DB）;
    在session_start（）;

Here is the PHP which I think is ok just including this for completeness: $connect = mysql_connect("$mysql_host", "$mysql_user", "$mysql_password")or die("cannot connect"); mysql_select_db("$mysql_database", $connect)or die("cannot select DB"); session_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"];
}

任何帮助，这将是非常美联社preciated，试图让我的头围绕这一切的HTTP发布的东西。

Any help with this would be very appreciated, trying to get my head around all this HTTP POSTing stuff.

推荐答案

发送一个HTTP请求和读回HTTP响应的例子：

Example of sending an HTTP request and reading back the HTTP response:

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);

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