Yii2不是唯一的表/别名： '用户'

Question

在 “门票” 的模式：

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


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

在 “TicketSearch” 模式：

$query->joinWith(['user','supervisor']); 

    $query 
     ->andFilterWhere(['like', 'user.surname', $this->user_id]) 
     ->andFilterWhere(['like', 'user.surname', $this->supervisor_id] 

如果我尝试搜索主管，得到这个错误：

Not unique table/alias: 'user' 
The SQL being executed was: SELECT COUNT(*) FROM `ticket` LEFT JOIN `user` ON `ticket`.`user_id` = `user`.`id` LEFT JOIN `user` ON `ticket`.`supervisor_id` = `user`.`id` WHERE `user`.`surname` LIKE '%surname4%' 

我曾试图改变名称的用户表：

public function getSupervisor() 
    { 
     return $this->hasOne(User::className(), ['id' => 'supervisor_id']) 
->from(User::tableName() . 'u2'); 
    } 

但这个错误返回：

You have an error in your SQL syntax; check the manual that corresponds 
to your MySQL server version for the right syntax to use near 'u2.`id` 
WHERE `supervisor`.`surname` LIKE '%surname4%'' at line 1 The SQL being executed was: 
SELECT COUNT(*) FROM `ticket` LEFT JOIN `user` ON `ticket`.`user_id` = `user`.`id` LEFT JOIN `user`u2 ON 
`ticket`.`supervisor_id` = `user`u2.`id` WHERE `supervisor`.`surname` LIKE '%surname4%' 

Answer 1

您别名之前错过了空间。应该是：

public function getSupervisor() 
{ 
    return $this->hasOne(User::className(), ['id' => 'supervisor_id']) 
     ->from(User::tableName() . ' u2'); 
} 

你也可以将它指定为数组键：

public function getSupervisor() 
{ 
    return $this->hasOne(User::className(), ['id' => 'supervisor_id']) 
     ->from(['u2' => User::tableName()]); 
} 

甚至joinWith()与此关系：

->joinWith([ 
    'supervisor' => function ($query) { 
     /* @var $query \yii\db\ActiveQuery */ 

     $query->from(User::tableName() . ' u2'); 
     // or $query->from(['u2' => User::tableName()]); 
    }, 
]); 

正式文件：

• from()