教义:如何从一个实体穿越到另一个“链接”实体？

Question

我正在使用cross-join在Doctrine_RawSql中加载3个不同的表。这使我返回以下对象：

User              -> User class (doctrine base class)
    Settings      -> DoctrineCollection of Setting
    User_Settings -> DoctrineCollection of User_Setting

上面的对象是many-to-many关系User和Setting之间的结果，其中User_Setting充当引用表。User_Setting还包含另一个名为value的字段。这显然包含了相应Setting的值。

但是，到目前为止，返回的Settings对象的User_Settings 属性在中是没有办法相互链接的，(当然除了setting_id字段之外)。

是否有任何直接方式直接从Settings User_Settings 属性遍历到相应的属性？

这是相应的查询：

        $sets = new Doctrine_RawSql();
        $sets->select('{us.*}, {s.*}, {uset.*}')
        ->from('(User us CROSS JOIN Setting s) LEFT JOIN User_Setting uset ON us.user_id = uset.user_id AND s.setting_id = uset.setting_id')
        ->addComponent('us', 'User us')
        ->addComponent('uset', 'us.User_Setting uset')
        ->addComponent('s', 'us.Setting s')
        ->where('s.category_id = ? AND us.user_id = ?',array(1, 1));

        $sets = $sets->execute();

编辑：

1:这是相关的YAML标记

//User relations:
Setting:
  class: Setting
  foreignAlias: User
  refClass: User_Setting
  local: user_id
  foreign: setting_id

//Setting relations:
User:
  class: User
  foreignAlias: Setting
  refClass: User_Setting
  local: setting_id
  foreign: user_id

//User_Setting relations:
Setting:
  foreignAlias: User_Setting
  local: setting_id
  foreign: setting_id
User:
  foreignAlias: User_Setting
  local: user_id
  foreign: user_id  

2.这是目标代码(由YAML生成)：

//BaseUser setup()
    $this->hasMany('Setting', array(
         'refClass' => 'User_Setting',
         'local' => 'user_id',
         'foreign' => 'setting_id'));
    $this->hasMany('User_Setting', array(
         'local' => 'user_id',
         'foreign' => 'user_id'));

//BaseSetting setup()
    $this->hasMany('User', array(
         'refClass' => 'User_Setting',
         'local' => 'setting_id',
         'foreign' => 'user_id'));
    $this->hasMany('User_Setting', array(
         'local' => 'setting_id',
         'foreign' => 'setting_id'));

//BaseUser_Setting setup()
    $this->hasOne('Setting', array(
         'local' => 'setting_id',
         'foreign' => 'setting_id'));
    $this->hasOne('User', array(
         'local' => 'user_id',
         'foreign' => 'user_id'));

Answer 1

实际上，您应该在ref表中定义一对多的关系，但是使用hasOne()：

class User_Setting
{
....
  public function setUp()
  {
    parent::setUp();
    $this->hasOne('User', array(
         'local' => 'user_id',
         'foreign' => 'id'));

    $this->hasOne('Setting', array(
         'local' => 'setting_id',
         'foreign' => 'id'));
  }
}

然后，您将拥有一个Doctrine_Record而不是Doctrine_Collection。

Answer 2

您可以在User_Setting类中定义缺失的关系：

class User_Setting
{
    // ...
    public function setUp()
    {
        $this->hasMany('Users', array(
            'class' => 'User',
            'local' => 'user_id',
            'foreign' => 'id',
        ));
        $this->hasMany('Settings', array(
            'class' => 'Setting',
            'local' => 'setting_id',
            'foreign' => 'id',
        ));
    }
}