1

我有一个有一些关系的数据库。在这个例子中我有两个表：

activity_ingredients   ingredient_aliases
---------------        ---------------
id                     id
ingredient_id          ingredient_id

我想在我的模型中设置一个条件，以通过字段“ingredient_id”连接表。我已经在这种模式下完成了：

class ActivityIngredients extends AppModel{
        public $name = 'ActivityIngredients'; 
        public $useTable = 'activity_ingredients';

        public $hasMany = array (
        'IngredientAlias' => array (
            'className'     => 'IngredientAlias',
            'foreignKey'   => false,
            'conditions' => array('ActivityIngredients.ingredient_id = IngredientAlias.ingredient_id')
        )
        );
    }   

class IngredientAlias extends AppModel {

    public $name = 'IngredientAlias';
    public $useTable = 'ingredient_aliases';
    public $belongsTo = array(
         'ActivityIngredients' => array(
            'className'    => 'ActivityIngredients',
            'foreignKey'   => false
        )

    );  
}

如果我写这段代码给我一个这样的错误：

未找到列：1054“where 子句”中的未知列“ActivityIngredients.ingredient_id”

但是，如果我写 hasOne 而不是 hasMany，它不会返回任何错误并且是完美的。但是我的关系是hasMany..为什么？

为了检索数据，我对我的 UserController 进行了查询。用户通过hasMany与Activity关联，Activity与ActivityIngredients在这种模式下关联：

public $hasOne = array(
            'ActivityIngredients' => array(
                'className' => 'ActivityIngredients',
                'conditions' => '',
                'dependent' => true,
                'foreignKey'   => 'activity_id'
            )

要检索数据，我目前正在使用此查询，但我认为我现在必须更改所有查询

$this->User->Activity->recursive = 2;
        $activity = $this->User->Activity->findAllByUser_id($id);
        $this->set('activity', $activity);
4

2 回答 2

0

在定义模型类时，首先应该使用单数词。这意味着模型名称应该是ActivityIngredient而不是ActivityIngredients。

虽然如果在activity_ingredients表中如果ingredient_idid 是属于ingredient_aliases表的外键。那么它应该被命名为ingredient_alias_id. 同样，如果您在ingredient_aliases表中执行相同操作，即外键名称将是activity_ingredient_id. 那么就完全没有问题了。

但是，我正在为您的问题提供解决方案。请检查和验证。

然后您的代码应如下所示：

class ActivityIngredient extends AppModel{
    public $name = 'ActivityIngredient'; 
    public $useTable = 'activity_ingredients'; // if you use singular term then no need to define $useTable

    public $hasMany = array (
    'IngredientAlias' => array (
        'className'     => 'IngredientAlias',
        'foreignKey'   => false,
        'finderQuery' => 'select * from ingredient_aliases as `IngredientAlias`
                          where `IngredientAlias`.`ingredient_id` = {$__cakeID__$}'
                               )
    );
}   

class IngredientAlias extends AppModel {

    public $name = 'IngredientAlias';
    public $useTable = 'ingredient_aliases';
    public $belongsTo = array(
                      'ActivityIngredient' => array(
                                   'className'    => 'ActivityIngredient',
                                   'foreignKey'   => 'ingredient_id' 
                                                   )
                             );  
}
于 2012-07-25T04:12:02.537 回答
0

你用过

条件' => 数组（'ActivityIngredients.ingredient_id = IngredientAlias.ingredient_id'）

它应该是

条件' => array('ActivityIngredients.ingredient_id => IngredientAlias.ingredient_id')

使用下面的代码

            public $hasMany = array (
            'IngredientAlias' => array (
                'className'     => 'IngredientAlias',
                'foreignKey'   => false,
                conditions' => array('ActivityIngredients.ingredient_id => IngredientAlias.ingredient_id')
            )
            );
        }   

    class IngredientAlias extends AppModel {

        public $name = 'IngredientAlias';
        public $useTable = 'ingredient_aliases';
        public $belongsTo = array(
             'ActivityIngredients' => array(
                'className'    => 'ActivityIngredients',
                'foreignKey'   => false
            )

        );  
    }
    ?>
于 2012-07-25T04:34:29.367 回答