Question

I am using CakePHP 2.0 and I have a Model with this ‘virtualFields’:

Country.php:

var $virtualFields = array(
    'path' => "CONCAT_WS('/', dirname, basename)"
);

If I do this in a Controller which uses “User”

$users = $this->User->find('all');

the virtual field path is set.

If I use this in my Controller, which also uses “User”

$options['fields'] = array(
    'DISTINCT User.*'
);
$options['joins'] = array(
    array(
        'table' => 'courses',
        'type' => 'inner',
        'conditions' => array(
            'User.id = courses.user_id'
        )
    ),
    array(
        'table' => 'times',
        'type' => 'inner',
        'conditions' => array(
            'courses.id = times.course_id'
        )
    )
);
$options['conditions'] = array(
    'times.amount > ' => 0
);

$users = $this->User->find('all', $options);

With the options, the path field is not set, of course, the SQL query does not seem to have the “CONCAT_WS(‘/’, dirname, basename)” field included, which is included if I do the find operation without the options.

What can I do having the options, so that the virtual field is automatically included? Of course I can write in the fields options the CONCAT, but that’s not very nice, especially if I change it.

Best regards.

Answer 1

Use group by rather than distinct and remove the fields option completely.

$options['group'] = array(
    'User.*'
);
$options['joins'] = array(
    array(
        'table' => 'courses',
        'type' => 'inner',
        'conditions' => array(
            'User.id = courses.user_id'
        )
    ),
    array(
        'table' => 'times',
        'type' => 'inner',
        'conditions' => array(
            'courses.id = times.course_id'
        )
    )
);
$options['conditions'] = array(
    'times.amount > ' => 0
);

$users = $this->User->find('all', $options);