我有以下型号:
User
列 {id,user_name,password,user_type}
Admin
列 {id,user_id,full_name,.....etc}
Editor
列 {id, user_id,full_name,...等}
关系是
User
:'admin' => array(self::HAS_ONE, 'Admin', 'user_id'),'editor' => array(self::HAS_ONE, 'Editor', 'user_id'),
Admin
:'user' => array(self::BELONGS_TO, 'User', 'user_id'),
Editor
:'user' => array(self::BELONGS_TO, 'User', 'user_id'),
fullName
现在我在模型中设置了一个虚拟属性,User
如下所示
public function getFullName()
{
if($this->user_type=='admin')
return $this->admin->full_name;
else if($this->user_type=='editor')
return $this->editor->full_name;
}
我可以在 gridview 中显示虚拟属性 , fullName
,但是如何向属性添加过滤器并使其在 gridview 中可排序?
更新1:
我根据@Jon 的回答更新了模型 search() 函数,如下所示
public function search()
{
$criteria=new CDbCriteria;
$criteria->select=array('*','COALESCE( editor.full_name,admin.first_name, \'\') AS calculatedName');
$criteria->with=array('editor','admin');
$criteria->compare('calculatedName',$this->calculatedName,true);
$criteria->compare('email',$this->email,true);
$criteria->compare('user_type',$this->user_type);
return new CActiveDataProvider($this, array(
'criteria'=>$criteria,
));
}
管理员和编辑者的姓名都正确显示在网格视图中。但是当我通过过滤器进行搜索时,会发生以下异常,
CDbCommand failed to execute the SQL statement: SQLSTATE[42S22]: Column not found: 1054 Unknown column 'calculatedName' in 'where clause'. The SQL statement executed was: SELECT COUNT(DISTINCT `t`.`id`) FROM `user` `t` LEFT OUTER JOIN `editor` `editor` ON (`editor`.`user_id`=`t`.`id`) LEFT OUTER JOIN `admin` `admin` ON (`admin`.`user_id`=`t`.`id`) WHERE (calculatedName LIKE :ycp0) (C:\xampplite\htdocs\yii\framework\db\CDbCommand.php:528)</p><pre>#0 C:\xampplite\htdocs\yii\framework\db\CDbCommand.php(425):
我怎样才能摆脱这个?
更新2:我的错误。当我改变线路时它工作正常
$criteria->compare('calculatedName',$this->calculatedName,true);
至
$criteria->compare('COALESCE( editor.full_name,admin.first_name, \'\')',$this->calculatedName,true);
顺便说一句,谢谢@Jon。