Question

我有一个正常运行的Laravel Query Builder代码段：

$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);

有没有办法避免使用原始查询并获得相同的结果？换句话说，我怎么能用更“查询构建器”的风格来写这个呢？我还可以将其转换为Eloquent查询吗？

由于

Answer 1

您可以使用selectSub方法进行查询。

（1）首先创建角色查询

$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）第二次将$role子查询添加为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);

输出SQL语法

"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"

Laravel：在“query-builder”或“eloquent”中更改原始查询

1 个答案: