PHP FLEX错误:在连接和验证服务错误时:致命错误:只能通过引用传递变量

我收到错误:致命错误:只有变量可以通过引用传递 当我尝试在flex builder中验证服务时 (!)致命错误:只有变量可以在“phpfile”中通过引用传递 这就是我打电话的时候:
mysqli_stmt_bind_result($stmt, $row->city.Id, $row->prov.Id, $row->dist.Id, $row->coun.Id);
这里是整个文件:
<?php
class TblcityService {
 var $username = "root";
 var $password = "";
 var $server = "localhost";
 var $port = "3306";
 var $databasename = "xoffercommon";
 var $tablename = "tblcity";
 var $connection;
 public function __construct() {
    $this->connection = mysqli_connect(
  $this->server,  
  $this->username,  
  $this->password, 
  $this->databasename,
  $this->port
  );
  mysqli_set_charset($this->connection,'utf8');

 $this->throwExceptionOnError($this->connection);
}

public function getCityProvDistCoun($cityID) {
$query = "SELECT city.Id, prov.Id, dist.Id, coun.Id FROM ((tblCity AS city INNER JOIN tblProvence AS prov ON city.tblProvence_Id = prov.ID) INNER JOIN tblDistrict AS dist ON prov.tblDistrict_Id = dist.ID) INNER JOIN tblCountry AS coun ON dist.tblCountry_Id = coun.ID WHERE city.id = ?";
$stmt = mysqli_prepare($this->connection, $query);
$this->throwExceptionOnError();

mysqli_stmt_bind_param($stmt, 'i', $cityID);  
$this->throwExceptionOnError();

mysqli_stmt_execute($stmt);
$this->throwExceptionOnError();

mysqli_stmt_bind_result($stmt, $row->city.Id, $row->prov.Id, $row->dist.Id, $row->coun.Id);

if(mysqli_stmt_fetch($stmt)) {
      return city.Id;
} else {
      return null;
}
}
/**
 * Utility function to throw an exception if an error occurs 
 * while running a mysql command.
 */
private function throwExceptionOnError($link = null) {
 if($link == null) {
  $link = $this->connection;
 }
 if(mysqli_error($link)) {
  $msg = mysqli_errno($link) . ": " . mysqli_error($link);
  throw new Exception('MySQL Error - '. $msg);
 }  
}
}
?>
与我用php运行的其余查询相比,我看到的唯一区别是,这次它是一个联合并使用“tablename”。“fieldname”表示法,可能是“。”给出参考错误?我是这样,什么是sybt     
已邀请:
return city.Id;
应该是
return row->city.Id;
    
php中的点用于连接字符串。 $ row-> city.Id也不应该是你所期望的。     

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