Question

我有两个名为

的表

new_tbl_master_siteid - ＆gt;型号：SiteID
app_tbl_ict_dapot_genset - ＆gt;型号：发电机组

我有一个像这样的mysql查询

SELECT A.siteid, B.type, B.status_operations, B.status
FROM new_tbl_master_siteid as A
LEFT JOIN app_tbl_ict_dapot_genset as B
ON A.siteid = B.siteid

我想离开的地方加入 siteid 列

我正在考虑修改genset模型上的CdbCriteria，我该如何做到这一点，因此genset gridview的数据提供者也可以使用SiteID模型中的列？

发电机组模型的标准

$criteria=new CDbCriteria;

        $criteria->compare('id',$this->id);
        $criteria->compare('siteid',$this->siteid,true);
        $criteria->compare('merk',$this->merk,true);
        $criteria->compare('type',$this->type,true);
        $criteria->compare('capacity',$this->capacity,true);
        $criteria->compare('kapasitas_tangki',$this->kapasitas_tangki,true);
        $criteria->compare('status',$this->status,true);
        $criteria->compare('status_operations',$this->status_operations,true);

        return new CActiveDataProvider($this, array(
            'criteria'=>$criteria,

==================

以下是

class genset extends CActiveRecord
.....
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(
            'siteids'=>array(self::BELONGS_TO,'SiteID','siteid'),
        );
    }
....

public function search()
    {
        // @todo Please modify the following code to remove attributes that should not be searched.

        $criteria=new CDbCriteria;

        $criteria->compare('id',$this->id);
        $criteria->compare('siteid',$this->siteid,true);
        $criteria->compare('merk',$this->merk,true);
        $criteria->compare('type',$this->type,true);
        $criteria->compare('capacity',$this->capacity,true);
        $criteria->compare('kapasitas_tangki',$this->kapasitas_tangki,true);
        $criteria->compare('status',$this->status,true);
        $criteria->compare('status_operations',$this->status_operations,true);
        $criteria->with = array('siteids.rtp');
        $criteria->compare('siteids.rtp',$this->rtp,true);

        return new CActiveDataProvider($this, array(
            'criteria'=>$criteria,
        ));
    }

和siteid类

class SiteID extends CActiveRecord
.....
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(
            'gensets'=>array(self::HAS_MANY, 'genset', 'siteid'),   
        );
    }

但我收到了错误

Relation "rtp" is not defined in active record class "SiteID".

rtp是siteid模型中的一列

Answer 1

Yii这样做的方法是通过关系，例如你应该将连接表的关系添加到你的模型中（如果它还没有）然后你可以直接访问相关模型的属性，例如，如果SiteID对象是$ model和关系的名称是'genset'，你可以得到genset的状态：$ model-＆gt; genset-＆gt; status

Answer 2

你要做的第一件事是在模型之间建立关系，可以这样做：

public function relations()
{
        return array(
            'genset'=>array(self::HAS_MANY, 'genset', 'siteid'),
        );
}

在模型类中。在两个模型中定义关系后，修改您的条件以使用->with

加入结果

$criteria->with = array('genset');

你准备好了。但请确保在两个模型中定义关系。

将MySQL查询转换为Yii CDbCriteria

2 个答案: