Laravel - 使用join和concat的Querybuilder

MDS*_*MDS 6 php mysql query-builder eloquent laravel-4

我试图从users表中提取与users_groups数据透视表中某个组匹配的所有用户.我正在使用Cartalyst btw的Sentry 2.

这可以使所有用户连接名字和姓氏.

User::select(DB::raw('CONCAT(last_name, ", ", first_name) AS full_name'), 'id')
        ->where('activated', '=', '1')
        ->orderBy('last_name')
        ->lists('full_name', 'id');
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当我尝试将其更改为也过滤不属于某个组的用户时,我收到语法错误.

User::select(DB::raw('SELECT CONCAT(user.last_name, ", ", user.first_name) AS user.full_name'), 'user.id', 'users_groups.group_id', 'users_groups.user_id')
                        ->join('users_groups', 'user.id', '=', 'users_groups.user_id')
                        ->where('user.activated', '=', '1')
                        ->where('users_groups.group_id', '=', $group)
                        ->orderBy('user.last_name')
                        ->lists('user.full_name', 'user.id');
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任何朝着正确方向的推动都将非常感激.

编辑:语法错误

SQLSTATE[42000]: Syntax error or access violation: 1064 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 'SELECT CONCAT(user.last_name, ", ", 
user.first_name) AS user.full_name, `user`.`' at line 1 (SQL: select SELECT 
CONCAT(user.last_name, ", ", user.first_name) AS user.full_name, `user`.`id`, 
`users_groups`.`group_id`, `users_groups`.`user_id` from `users` inner join 
`users_groups` on `user`.`id` = `users_groups`.`user_id` where 
`users`.`deleted_at` is null and `user`.`activated` = 1 and 
`users_groups`.`group_id` = 9 order by `user`.`last_name` asc)
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MDS*_*MDS 13

洛根的回答让我开始朝着正确的方向前进.我还必须删除所有'用户'.我猜想,因为它正在调用用户模型.此查询有效:

User::select(DB::raw('CONCAT(last_name, ", ", first_name) AS full_name'), 'id')
                        ->join('users_groups', 'id', '=', 'users_groups.user_id')
                        ->where('activated', '=', '1')
                        ->where('users_groups.group_id', '=', $group)
                        ->orderBy('last_name')
                        ->lists('full_name', 'id');
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感谢大家!希望如果其他人碰到这个,他们会找到这个问题的指导.


Maj*_*bib 10

您不需要在DB :: raw()中选择,请参阅示例

User::select(
          'id',
          DB::raw('CONCAT(first_name," ",last_name) as full_name')

        )
       ->orderBy('full_name')
       ->lists('full_name', 'id');
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