我的数据库中有带有“服务器”名称的表。因此,当我想使用该代码添加一些数据时,就可以了:
public function action_add()
{
$serverGameId = (int)Arr::get($_POST, 'serverGameId', '');
$newServerName = Arr::get($_POST, 'newServerName', '');
try
{
$server = new Model_Server();
$server->Name = $newServerName;
$server->GameId = $serverGameId;
$server->save();
}
catch(ORM_Validation_Exception $e)
{
echo json_encode(array("success" => false, "errors" => $e->errors('validation')));
return false;
}
$this->request->headers['Content-Type'] = 'application/json';
echo json_encode(array("success" => true, "serverId" => $server->Id));
return true;
}
这是一个模型:
class Model_Server extends ORM {
protected $_table_name = 'servers';
public function rules()
{
return array(
'Name' => array(
array('not_empty'),
)
);
}
}
但是当我尝试从表中选择它时遇到问题:
public function action_servers()
{
$gameId = (int)Arr::get($_POST, 'gameId', '');
if($gameId == -1) return false;
try
{
$servers = ORM::factory('server')
->where('GameId', '=', $gameId)
->find_all();
}
catch(ORM_Validation_Exception $e)
{
echo json_encode(array("success" => false, "errors" => $e->errors('validation')));
return false;
}
$this->request->headers['Content-Type'] = 'application/json';
echo json_encode(array("success" => true, "servers" => $servers, "gameId" => $gameId));
return true;
}
我已经尝试通过在 try 块中更改代码来解决问题:
$servers = DB::select('servers')->where('GameId', '=', $gameId);
即使我尝试在没有'-> where'的情况下从数据库中获取所有服务器,它也不起作用。
有任何想法吗?