从 PHP 文件接收 HTTP POST 回显响应（发送 POSTS 工作正常，这是我无法弄清楚的接收）

Posted 2023-02-25

【中文标题】从 PHP 文件接收 HTTP POST 回显响应（发送 POSTS 工作正常，这是我无法弄清楚的接收）

【英文标题】：recieving HTTP POST echo response from a PHP file (sending the POSTS works fine, it's the receive that I can't figure out)

【发布时间】：2012-08-11 09:17:57

【问题描述】：

所以正如标题所暗示的那样，我的问题是得到对我正在制作的 HTTP POST 的响应。 应该发生的是我发送了一堆变量，php 检查数据库并将结果发回给我（作为页面的回显）。

这里是安卓代码：

 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，我认为只是为了完整起见包含这个是可以的： $connect = mysql_connect("$mysql_host", "$mysql_user", "$mysql_password") or die("无法连接"); mysql_select_db("$mysql_database", $connect)or die("不能选择数据库"); 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"];

非常感谢您对此提供的任何帮助，让我了解所有这些 HTTP POST 内容。

【参考方案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);

【讨论】：

因此，要将其融入到我已经在做的事情中，我可以在我已经拥有的 InputStream 上使用缓冲阅读器，这样可以吗？

@RaidenXL 是的。它在我的应用程序上对我有用。我相信您也可以使用getEntity() 选项，但我没有任何经验，因此无法为您提供帮助。