Question

尝试执行此查询：

SELECT md.title, md.url 
   FROM MenuData md, Menu m 
   WHERE m.parent=0 
      AND m.menuType=1 
      AND md.menuID = m.id 
  ORDER BY m.sort ASC

如下所示：

  $criteria = new CDbCriteria();
            $criteria->together = true;
            $criteria->with = array('MenuData');
            $criteria->compare("Menu.menuType", 1);
            $criteria->compare("Menu.parent", 0);
            $criteria->join = 'LEFT JOIN MenuData ON MenuData.menuID=Menu.id';
            $criteria->order = '`Menu.sort` ASC';
            $criteria->compare("MenuData.locale", $this->lang);
            $results = Menu::model()->findAll($criteria);

同时拥有Menu和Menudata模型。但是得到以下错误：

Relation "MenuData" is not defined in active record class "Menu".

Answer 1

$ criteria-＆gt; with == join。

我假设你的模特菜单中有关系。

 public function relations()
{
    // NOTE: you may need to adjust the relation name and the related
    // class name for the relations automatically generated below.
    return array(
        'MenuData' => array(self::HAS_MANY, 'Menudata', 'menuID','joinType' => 'LEFT JOIN'),
    );
}

您的标准：

  $criteria = new CDbCriteria();
        $criteria->together = true;
        $criteria->with = array('MenuData');
        $criteria->compare("t.menuType", 1);
        $criteria->compare("t.parent", 0);
       // $criteria->join = 'LEFT JOIN MenuData ON MenuData.menuID=Menu.id';
        $criteria->order = '`t.sort` ASC';
        $criteria->compare("MenuData.locale", $this->lang);
        $results = Menu::model()->findAll($criteria);

Yii：使用多个表格

1 个答案: