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
没有找到相关结果
已邀请:
2 个回复
僻朵庙惩竣
应该是
悲帽慑彤电