Laravel：在「查詢生成器」或「雄辯」一個

Question

我有這個Laravel查詢生成器片段，做工精細更改原始查詢：

$records = DB::table('users') 
    ->select(
     DB::raw('users.*, activations.id AS activation, 
       (SELECT roles.name FROM roles 
        INNER JOIN role_users 
        ON roles.id = role_users.role_id 
        WHERE users.id = role_users.user_id LIMIT 1) 
        AS role') 
    ) 
    ->leftJoin('activations', 'users.id', '=', 'activations.user_id') 
    ->where('users.id', '<>', 1) 
    ->orderBy('last_name') 
    ->orderBy('first_name') 
    ->paginate(10); 

有沒有辦法避免使用原始查詢並得到相同的結果？換句話說，我怎樣才能以更多的「查詢構建器」風格來編寫它？我也可以把它翻譯成一個Eloquent查詢嗎？

感謝

Answer 1

您可以使用您的查詢selectSub方法。

（1）首先創建角色查詢

$role = DB::table('roles') 
      ->select('roles.name') 
      ->join('roles_users', 'roles.id', '=', 'role_users.role_id') 
      ->whereRaw('users.id = role_users.user_id') 
      ->take(1); 

（2）第二添加$role子查詢作爲role

DB::table('users') 
       ->select('users.*', 'activations.id AS activation') 
       ->selectSub($role, 'role') // Role Sub Query As role 
       ->leftJoin('activations', 'users.id', '=', 'activations.user_id') 
       ->where('users.id', '<>', 1) 
       ->orderBy('last_name') 
       ->orderBy('first_name') 
       ->paginate(10); 

輸出SQL語法

"select `users`.*, `activations`.`id` as `activation`, 
(select `roles`.`name` from `roles` inner join `roles_users` on `roles`.`id` = `role_users`.`role_id` 
where users.id = role_users.user_id limit 1) as `role` 
from `users` 
left join `activations` on `users`.`id` = `activations`.`user_id` 
where `users`.`id` <> ? 
order by `last_name` asc, `first_name` asc 
limit 10 offset 0"