Yii Framework : Join table (or other SQL) in data provider? - php

Question

Background
I have used Gii Crud Generator with my "Category" model, and I want to modify the admin form.
I look inside "protected/views/Category/admin.php,
I found the table is render by a widget('zii.widgets.grid.CGridView'),
and it using a data Provider for it's data.
I suppose I can find some where to input the SQL query in the data Provider, but I don't understand about how's it works.
these is the code In the Model->relations(), but I don't know what to do next.
public function relations(){
return array(
'cateLang' => array(self::HAS_MANY, 'CategoryLang', 'cate_id')
);
}
where the data provider is generated :
public function search(){
$criteria=new CDbCriteria;
$criteria->compare('id',$this->id);
$criteria->compare('status',$this->status,true);
$criteria->compare('createDate',$this->createDate,true);
$criteria->compare('updateDate',$this->updateDate,true);
$criteria->compare('remark',$this->remark,true);
return new CActiveDataProvider($this->with('cateLang'), array(
'criteria'=>$criteria,
));
}
Target
I want to add two more columns at the table of "protected/views/Category/admin.php,
which will show French Title & English Title of the row.
To get data in SQL, it will be :
SELECT
cate.id,
lang1.name as "FrenchTitle",
lang2.name as "EnglishTitle",
cate.updateDate,
cate.createDate,
cate.remark
FROM `category` cate
LEFT JOIN `categorylang` lang1
ON `lang1`.`cate_id` = `cate`.id
AND `lang1`.`lang_id`= 1
LEFT JOIN `categorylang` lang2
ON `lang2`.`cate_id` = `cate`.id
AND `lang2`.`lang_id`= 2
WHERE cate.status = 'live'
If I can done with data Provider, the CGridView parameter may be like this :
$this->widget('zii.widgets.grid.CGridView', array(
'id'=>'category-grid',
'dataProvider'=>$model->search(),
'filter'=>$model,
'columns'=>array(
'id',
'FrenchTitle',
'EnglishTitle',
'createDate',
'updateDate',
'remark',
array(
'class'=>'CButtonColumn',
),
),
));

Answer 1

You could try the following:
public function search(){
$criteria=new CDbCriteria;
$criteria->compare('id',$this->id);
$criteria->compare('status',$this->status,true);
$criteria->compare('createDate',$this->createDate,true);
$criteria->compare('updateDate',$this->updateDate,true);
$criteria->compare('remark',$this->remark,true);
$criteria->with = array('cateLang' => array(
'condition' => 'cateLang.id = 1 OR cateLang.id = 2',
'order' => 'cateLang.id ASC'
));
return new CActiveDataProvider($this, array(
'criteria'=>$criteria,
));
}
$this->widget('zii.widgets.grid.CGridView', array(
'id'=>'category-grid',
'dataProvider'=>$model->search(),
'filter'=>$model,
'columns'=>array(
'id',
array(
'name' => 'FrenchTitle'
'value' => '(isset($data->cateLang[0])) ? $data->cateLang[0]->name : "no Title"',
),
array(
'name' => 'EnglishTitle'
'value' => '(isset($data->cateLang[1])) ? $data->cateLang[1]->name : "no Title"',
),
'createDate',
'updateDate',
'remark',
array(
'class'=>'CButtonColumn',
),
),
));
In the search I specify that I want only cateLang object with the id 1 or 2 and then in the cgridview I display a relational object.