从指向相同 table 的两个关系中过滤 first_name 和 last_name

Question

我在 user table 中有列 first_name 和 last_name，这又与 order_item table 相关 instructor_id 和 user_id 都链接到 user table.

现在我在 OrderItem 模型中定义了如下关系：

public function getInstructor() {
        return $this->hasOne(User::className(), ['id' => 'instructor_id']);
    }

和

public function getCustomerName() {
        return $this->hasOne(User::className(), ['id' => 'user_id']);
    }

并在 OrderSearch 模型中

public $instructor;
public $name;

$query->joinWith(['instructor','location1','customerName n']);        

$query->andFilterWhere(['like', 'concat(user.first_name," ",user.last_name)',  $this->name]) 
        ->$query->andFilterWhere(['like', 'concat(user.first_name," ",user.last_name)',  $this->instructor]);

现在我收到类似

的错误

Method yii\db\ActiveQuery::__toString() must not throw an exception, 
caught Exception: Serialization of 'Closure' is not allowed

如果我在 joinwith 中删除别名 n，我会收到如下错误：语法错误或访问冲突：1066 不唯一 table/alias：'user'

The SQL being executed was: SELECT COUNT(*) FROM `order_item` LEFT JOIN `user` ON
`order_item`.`instructor_id` = `user`.`id` LEFT JOIN `location` ON `order_item`.`location_id` =  
`location`.`id` LEFT JOIN `user` ON `order_item`.`user_id` = `user`.`id`

Answer 1

你可以试着做成类似这样的：

class Order {
...
  //first relation with table COMPANY
  public function getCustomer()
  {
      return $this->hasOne( Company::className(), ['id' => 'customerid'] ) //table COMPANY
                  ->andOnCondition( [ '==relationAlias==.type' => 'customer' ] );
  }

  //second relation with table COMPANY
  public function getSupplier()
  {
      return $this->hasOne( Company::className(), ['id' => 'supplierid'] ) //table COMPANY
                  ->andOnCondition( [ '==relationAlias==.type' => 'supplier' ] );
  }
...
}

然后我们可以写

$orders = Order::find()
                   ->alias( 'o' )
                   ->innerJoinWith('customer c')
                   ->innerJoinWith('supplier s');

构建查询

select o.* from order o
join company c on c.id = o.customerid and c.type = 'customer'
join company s on s.id = o.supplierid and s.type = 'supplier'

这个问题是在 github 上提出的，并提供了解决方案： https://github.com/yiisoft/yii2/issues/10883

Answer 2

比我想的简单多了，但是花了很多时间才解决。但是我还是不明白为什么题中的方法不行

就像我所做的那样，我将查询分开如下，并在查询中使用了别名。

现有

$query->andFilterWhere(['like', 'concat(user.first_name," ",user.last_name)',  $this->name]) 
->$query->andFilterWhere(['like', 'concat(user.first_name," ",user.last_name)',  $this->instructor]);

已更新

$query->andFilterWhere(['like', 'concat(n.first_name," ",n.last_name)',  $this->name]); 
$query->andFilterWhere(['like', 'concat(user.first_name," ",user.last_name)',  $this->instructor]);

它工作正常。

从指向相同 table 的两个关系中过滤 first_name 和 last_name

filter first_name and last_name from two relation pointing to same table

php

mysql

yii2