Php Web服务总是返回null值

我在玩我找到的代码。我只能让我的数据库返回空json对象。这是我的密码

public class TableTalkSplash extends Activity {
    private InputStream is;

    /** Called when the activity is first created. */
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);

      String result = "";
      InputStream is = null;
      //the year data to send
      ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
      nameValuePairs.add(new BasicNameValuePair("row_id","1"));

      //http post
      try{
              HttpClient httpclient = new DefaultHttpClient();
              HttpPost httppost = new HttpPost("http://mywebsite.com/myscript.php");
              httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
              HttpResponse response = httpclient.execute(httppost);
              HttpEntity entity = response.getEntity();
              is = entity.getContent();
      }catch(Exception e){
              Log.e("log_tag", "Error in http connection "+e.toString());
      }
      //convert response to string
      try{
              BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8);
              StringBuilder sb = new StringBuilder();
              String line = null;
              while ((line = reader.readLine()) != null) {
                      sb.append(line + "\n");
              }
              is.close();

              result=sb.toString();
              System.out.println(result); // this is printing "null" to the console
      }catch(Exception e){
              Log.e("log_tag", "Error converting result "+e.toString());
      }

      //parse json data
      try{
              JSONArray jArray = new JSONArray(result);
              for(int i=0;i<jArray.length();i++){
                      JSONObject json_data = jArray.getJSONObject(i);
                      Log.i("log_tag","id: "+json_data.getInt("id")+
                              ", name: "+json_data.getString("name")+
                              ", sex: "+json_data.getInt("sex")+
                              ", birthyear: "+json_data.getInt("birthyear")
                      );
              }

      }catch(JSONException e){
              Log.e("log_tag", "Error parsing data "+e.toString()); //exception is being caught here
      }
    }
}

公共类TableTalkSplash扩展活动{
私有输入流是；
/**在首次创建活动时调用*/
@凌驾
创建时的公共void（Bundle savedInstanceState）{
super.onCreate（savedInstanceState）；
setContentView（R.layout.main）；
字符串结果=”；
InputStream=null；
//要发送的年度数据
ArrayList nameValuePairs=新的ArrayList（）；
添加（新的BasicNameValuePair（“行id”，“1”））；
//http post
试一试{
HttpClient HttpClient=新的DefaultHttpClient（）；
HttpPost HttpPost=新的HttpPost（“http://mywebsite.com/myscript.php");
setEntity（新的UrlEncodedFormEntity（nameValuePairs））；
HttpResponse response=httpclient.execute（httppost）；
HttpEntity=response.getEntity（）；
is=entity.getContent（）；
}捕获（例外e）{
e（“Log_标记”，“http连接错误”+e.toString（））；
}
//将响应转换为字符串
试一试{
BufferedReader reader=新的BufferedReader（新的InputStreamReader（is，“iso-8859-1”），8；
StringBuilder sb=新的StringBuilder（）；
字符串行=null；
而（（line=reader.readLine（））！=null）{
sb.追加（第+行“\n”）；
}
is.close（）；
结果=sb.toString（）；
System.out.println（result）；//这是向控制台打印“null”
}捕获（例外e）{
Log.e（“Log_标记”，“错误转换结果”+e.toString（））；
}
//解析json数据
试一试{
JSONArray jArray=新JSONArray（结果）；
对于（int i=0；i我猜db连接或查询失败，请尝试以下方法以帮助调试：

<?php

error_reporting(-1);
ini_set("display_errors", 1);

mysql_connect("host","username","password") or die('Could not connect: ' . mysql_error());;
mysql_select_db("mydb");

$query = "SELECT * FROM dinner_info WHERE row_id ='".mysql_real_escape_string($_REQUEST['row_id'])."'"

$q=mysql_query($query);

if (!$q) {
    $message  = 'Invalid query: ' . mysql_error() . "\n";
    $message .= 'Whole query: ' . $query;
    die($message);
}

while($e=mysql_fetch_assoc($q))
        $output[]=$e;

print(json_encode($output));

mysql_close();

要在PHP中打开错误报告，请将以下内容放在顶部：
错误报告（-1）；
。谢谢，这是我的问题的一部分，我不知道如何调试脚本语言ha。我将添加它并查看它的说明。在感谢您向我展示如何调试PHP之后，我将其添加到PHP脚本的顶部！我能够解决我的问题。
<?php

error_reporting(-1);
ini_set("display_errors", 1);

mysql_connect("host","username","password") or die('Could not connect: ' . mysql_error());;
mysql_select_db("mydb");

$query = "SELECT * FROM dinner_info WHERE row_id ='".mysql_real_escape_string($_REQUEST['row_id'])."'"

$q=mysql_query($query);

if (!$q) {
    $message  = 'Invalid query: ' . mysql_error() . "\n";
    $message .= 'Whole query: ' . $query;
    die($message);
}

while($e=mysql_fetch_assoc($q))
        $output[]=$e;

print(json_encode($output));

mysql_close();