Question

I have a site developed in cakephp 2.0, I want to make a relation between two tables:

activity_ingredients

1   id  int(10) UNSIGNED    No  None    AUTO_INCREMENT  
2   type_id     tinyint(2)  No  None        
3   activity_id     int(11)     No  None        
4   ingredient_id   int(10)     No  None        
5   created     datetime        

actions

1   id  int(10) UNSIGNED    No  None    AUTO_INCREMENT  
2   type_id     tinyint(2)  No  None        
3   language    char(2)     No  None        
4   text    varchar(100)        No  None        
5   created     datetime    

I want to associate the two tables with the field “type_id”.
I have done in this mode into my code:

    class Action extends AppModel{
    public $name = 'Action'; //non utilizzata nel sito è il nome del modello alla fine per migliorare la compatibilità

    public $belongsTo = array(
        'ActivityIngredients' => array(
            'className'     => 'ActivityIngredients',
            'conditions'    => '',
            'order'         => '',
            'foreignKey'    => 'type_id'
        )
    );

}

class ActivityIngredients extends AppModel{
        public $name = 'ActivityIngredients'; //non utilizzata nel sito è il nome del modello alla fine per migliorare la compatibilità

        public $belongsTo = array(
            'Activity' => array(
                'className'     => 'Activity',
                'conditions'    => '',
                'order'         => '',
                'foreignKey'    => 'activity_id'
            ),
            'Ingredient' => array(
                'className'     => 'Ingredient',
                'conditions'    => '',
                'order'         => '',
                'foreignKey'    => 'ingredient_id'
            )
        );

        public $hasMany = array(
            'Action' => array(
                'className' => 'Action',
                'conditions' => '',
                'dependent' => true,
                'foreignKey'   => 'type_id',
                'associatedKey'   => 'type_id'
            )
        );
    }

It doesn’t retrieve the correct data.. It seems that It takes the id for the foreign key.
This is the view:

<?php foreach ($user['Activity'] as $activities) {
var_dump($activities);
?>
    <div class="line-cnt"><div class="line">
    </div>
</div>
<h2>Attività</h2>
<div class="table">
    <div>
        <div>Activity created</div><div><?php echo $activities['created']; ?>
        </div>
    </div>
    <div>
        <div>Actions text</div><div><?php echo $activities['Action']['text']; ?></div>
    </div>
    <div>
        <div>ActivityIngredient ingredient_id</div><div><?php echo $activities['ActivityIngredients']['ingredient_id']; ?></div>
    </div>
</div>
<?php
}
?>

The controller is a simple query with find all and recursive 3 into the User that is collegate with the tables

$this->User->recursive = 3;
        $user = $this->User->read();

        if (empty($username) || $username != $user['User']['username']) {
            $this->redirect(array ('action'=>'view',$id,$user['User']['username']));
        }

        $this->set('user', $user);

Help me please

Answer 1

The first thing if you are using an “id” field in your “activity_ingredients” table, then you should use it as a foreignKey in another table.

A foreign key is a field in a relational table that matches a candidate key of another table.

Even if you are trying to use type_id as foreign key in “actions” table, Then type_id must be unique in your activity_ingredients table, and if it is so then you can define your ActivityIngredient Model as:

class ActivityIngredients extends AppModel{
    public $primaryKey = 'type_id';
    public $name = 'ActivityIngredients'; //non utilizzata nel sito è il nome del modello alla fine per migliorare la compatibilità

    public $belongsTo = array(
        'Activity' => array(
            'className'     => 'Activity',
            'conditions'    => '',
            'order'         => '',
            'foreignKey'    => 'activity_id'
        ),
        'Ingredient' => array(
            'className'     => 'Ingredient',
            'conditions'    => '',
            'order'         => '',
            'foreignKey'    => 'ingredient_id'
        )
    );

    public $hasMany = array(
        'Action' => array(
            'className' => 'Action',
            'conditions' => '',
            'dependent' => true,
            'foreignKey'   => 'type_id',
            'associatedKey'   => 'type_id'
        )
    );
}

And your Action Model will remain the same. And hence you will be able to fetch the desired records.

And even if you are not agree to define “type_id” as foreign key in your table. Then this code will work extremely well with your situation.

class ActivityIngredients extends AppModel{
public $name = 'ActivityIngredients'; //non utilizzata nel sito è il nome del modello alla fine per migliorare la compatibilità

public $belongsTo = array(
    'Activity' => array(
        'className'     => 'Activity',
        'conditions'    => '',
        'order'         => '',
        'foreignKey'    => 'activity_id'
    ),
    'Ingredient' => array(
        'className'     => 'Ingredient',
        'conditions'    => '',
        'order'         => '',
        'foreignKey'    => 'ingredient_id'
    )
);

public $hasMany = array(
    'Action' => array(
        'className' => 'Action',
        'conditions' => '',
        'dependent' => true,
        'foreignKey'   => false,
        'finderQuery'   => 'select * from actions as `Action` where
                            `Action`.`type_id` = {$__cakeID__$} '
    )
);

}

I am sure this will give you the desired result. Kindly ask if it not worked for you.