mysql - 我可以在 PhalconPHP 上为两个表构建模型吗？

Question

我有一个 games 表和两个表来存储他们的分数，game_scores 和 thirdparty_game_scores 具有完全相同的结构，如下所示:

-------------------------GAMES-------------------------
slug | title | technology | (other non-relevant fields)
-------------------------------------------------------

----------------------GAME SCORES----------------------
 id (PK) | game_slug | user | team | score | timestamp
-------------------------------------------------------

----------------THIRDPARTY GAME SCORES-----------------
 id (PK) | game_slug | user | team | score | timestamp
-------------------------------------------------------

他们的模特喜欢:

class GameScores extends BaseGamesModel {

  public function initialize() {
    parent::initialize();
    $this->belongsTo( 'game_slug', 'Games', 'slug' );
    $this->skipAttributes(array('creation_time', 'last_modified'));
  }
}

我正在建立每周排名，游戏分数仅存储在 game_scores 上:

$ranking = GameScores::sum(
  array(
    "column" => "score",
    "conditions" => $conditions,
    "group" => "team",
    "order" => "sumatory DESC"
  )
);

有没有什么方法可以创建一个模型，将来自 game_scores 和 thirdparty_game_scores 的所有数据串联起来，无需任何额外努力即可创建相同的排名？

非常感谢!

Answer 1

您可以在模型中动态设置源表。参见 Can I build a Model for two Tables on PhalconPHP?

您应该使用 setSource() 方法并将您的条件放在那里，例如:

   $source = 'default_table';
   if (class_name($this) === 'A') {
      $source = 'A_table'; 
   }
   return $source;