Yii 2:与同一个表的多重关系

Question

我遇到了Yii 2关系表的问题.我的工作有很多关系,但只有在这种情况下才会给我一个错误:

SQLSTATE [42S22]:找不到列:1054'where子句'中的未知列'father.name'

我认为问题是与同一个表"代理"的双重关系.查看模型中的代码段:

public function getAgent()
{
    return $this->hasOne(Agent::className(), ['id' => 'id_agent']);
}
public function getFather()
{
    return $this->hasOne(Agent::className(), ['id' => 'id_father']);
}

在我的GridView中,我看到了正确的值,但是当我尝试使用ORDER或"andWhere"进行过滤时,Yii2会返回错误.

您可以在下面找到searchModel的代码:

$dataProvider->sort->attributes['agentName'] = [ 
        'asc' => ['agent.name' => SORT_ASC],
        'desc' => ['agent.name' => SORT_DESC],
        'default' => SORT_ASC
    ];

$dataProvider->sort->attributes['fatherName'] = [
        //'asc' => ['father.name' => SORT_ASC],
        //'desc' => ['father.name' => SORT_DESC],
        'default' => SORT_ASC
    ];
//.......
$query->andFilterWhere(['like', 'agent.name', $this->agentName]);
$query->andFilterWhere(['like', 'father.name', $this->fatherName]);

agentName属性工作正常.有什么建议吗？谢谢!

-------更新:更多代码--------- searchModel:

public function search($params)
{
    $agent_aux = new Agent();
    $agent_id= $agent_aux->getAgentIdFromUser();

    if (Yii::$app->user->can('admin')){
        $query = Contract::find();
    }
    else{

        $query = Contract::find()->where(['contract.agent_id' => $agent_id]);            
    }

    $query->joinWith(['agent','seminar']);

    $dataProvider = new ActiveDataProvider([
        'query' => $query,
    ]);
    $this->load($params);
    $dataProvider->sort->attributes['seminar_location'] = [
        'asc' => ['seminar.location' => SORT_ASC],
        'desc' => ['seminar.location' => SORT_DESC],
    ];
    $dataProvider->sort->attributes['agentName'] = [ 
        'asc' => ['agent.name' => SORT_ASC],
        'desc' => ['agent.name' => SORT_DESC],
        'default' => SORT_ASC
    ];

    $dataProvider->sort->attributes['fatherName'] = [
        //'asc' => ['father.name' => SORT_ASC],
        //'desc' => ['father.name' => SORT_DESC],
        'default' => SORT_ASC
    ];
    if (!$this->validate()) {
        return $dataProvider;
    }
    $query->andFilterWhere([
        'id' => $this->id,
        'data' => $this->data,
        'id_agent' => $this->id_agent,
        'id_father' => $this->id_father,
        'id_seminar' => $this->id_seminar,
    ]);
    $query->andFilterWhere(['like', 'agent.name', $this->agentName]);
    $query->andFilterWhere(['like', 'father.name', $this->fatherName]);
    return $dataProvider;
}

Answer 1

您需要在模型中进行以下更改.from子句实际上是在创建一个别名.代理人和父亲关系将在单独的连接条款中被选中.在过滤条件中使用"代理"和"父"别名与列名称.

public function getAgent()
{
    return $this->hasOne(Agent::className(), ['id' => 'id_agent'])->from(['agent' => Agent::tableName()]);
}

public function getFather()
{
    return $this->hasOne(Agent::className(), ['id' => 'id_father'])->from(['father' => Agent::tableName()])
}

要改变的另一件事是

$query->joinWith(['agent','seminar', 'father']);