Question

我有一个关系数据库。我在这个例子中有两个表：

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

我想在我的模型中创建一个条件，通过字段“ingredient_id”连接表。我已经完成了这种模式：

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，请给我一个错误：

找不到列：1054'where子句'中的未知列'ActivityIngredients.ingredient_id'

但是如果不是hasMany我写了hasOne它不会返回任何错误并且是完美的。但我的关系是很多......为什么？

要检索数据，我会查询我的UserController。用户与hasMany的Activity相关联，Activity在此模式下与ActivityIngredients相关联：

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

要检索数据我现在正在使用此查询，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);

Answer 1

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为ActivityIngredient而不是ActivityIngredients。

虽然activity_ingredients表中的如果ingredient_id id是属于ingredient_aliases表的外键。然后它应该命名为ingredient_alias_id。同样，如果你在ingredient_aliases表中也这样做，即外键名称为activity_ingredient_id。那么根本就没有问题。

但是，我正在解决您的问题。请检查并验证。

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' 
                                                   )
                             );  
}

Answer 2

你已经用过了

conditions'=＆gt;数组（'ActivityIngredients.ingredient_id = IngredientAlias.ingredient_id'）

应该是

conditions'=＆gt; array（'ActivityIngredients.ingredient_id =＆gt; IngredientAlias.ingredient_id'）

使用以下代码

            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
            )

        );  
    }
    ?>

cakephp hasmany relationship给出错误

2 个答案: