Question

我有一个简单的查询，我应该以雄辩的方式运行，如下所示

 SELECT `followers`.*, `fol`.`followed_id` AS `is_fan` 
    FROM `followers` 
    LEFT JOIN `followers` AS `fol` ON `fol`.`follower_id` = `followers`.`followed_id` 
         AND fol.followed_id = ?
    WHERE `followers`.`followed_type` = ?    
    AND `followers`.`followed_id` = ? 
    GROUP BY `followers`.`followed_id`, `fol`.`followed_id`  
    ORDER BY `created_at` DESC

然后我用eloquent做了一个正确的查询：

$this->follow->with('follower')
             ->where('followers.followed_type', $followed['follower_type'])
             ->where('followers.followed_id',$followed['follower_id'])
             ->leftJoin('followers as fol',function($join) use ($followed)
             {
                $join->on('fol.follower_id','=','followers.followed_id')
                      ->where('fol.followed_id','=','followers.follower_id');
             })
             ->groupBy('followers.followed_id')
             ->groupBy('fol.followed_id')
             ->select('followers.*',"fol.followed_id as is_fan")
             ->get();

经过大量的调试以了解为什么列is_fan没有被填充为例外，我试图对where上传递的值进行硬编码在连接闭包中查看响应像这样：

->leftJoin('followers as fol',function($join) use ($followed)
                 {
                    $join->on('fol.follower_id','=','followers.followed_id')
                          ->where('fol.followed_id','=',1);
                 })

它已经工作了，所以我得到的解决方案是第三个参数不能是这个案例followers中父表的参数。有没有可能获得父表e的值作为where闭包中LeftJoin方法中的第三个参数传递？

Answer 1

您的代码唯一的问题是，where使用参数绑定，因此'followers.follower_id'在此处理为字符串：

->where('fol.followed_id','=','followers.follower_id');

DB::raw()在这里还不够，因为JoinClause s where的工作有点偏差，显然你只需要另一个on：

->on('fol.followed_id','=','followers.follower_id');

左边加入Eloquent，给父表的参数

1 个答案: