与HasMany的Yii Relation数据问题

Question

Currently we have two tables. Locations and Addresses

Locations are a 1 to many relation to addresses on addresses.location_id.

We are trying to output all addresses related to a specific location in the cgridview:

admin.php

<?php $this->widget('zii.widgets.grid.CGridView', array(
    'id'=>'locations-grid',
    'dataProvider'=>$location->search(),
    'filter'=>$location,
    'columns'=>array(
        'id',
        'name',
        array(
            'name' => 'address',
            'value' => 'implode("<br>",CHtml::listData($data->addresses(),"id","address"))',
            'filter' => CHtml::activeTextField($location, 'address'),
            'type' => 'html',
        ),/**/  
        array(
            'class'=>'CButtonColumn',
),
),
)); ?>

model (locations.php)

public function search()
{
        // Warning: Please modify the following code to remove attributes that
        // should not be searched.
        $criteria=new CDbCriteria;
        $criteria->together = true;
        $criteria->with = array('addresses');
        $criteria->compare('id',$this->id);
        $criteria->compare('name',$this->name,true);
        $criteria->compare('addresses.address',$this->address,true);
 
        return new CActiveDataProvider($this, array(
            'criteria'=>$criteria,
));
}

Is there a better way of going about getting the address data rather then imploding the $data->addresses function?

Answer 1

You can add a getter method to your Locations model.

public function getAddressNames($separator='<br>')
{
    $names = array();
    foreach($this->adresses as $address) {
$names[] = $address->address;
}
    return implode($separator, $names);
}

Then you can use addressNames like a regular attribute in your gridview.