Laravel 6，MYSQL-如何使用Laravel Querybuilder或Model Eloquent离开GroupBY的子查询？

问题描述

我尝试使用Laravel Eloquent来工作，但是我无法获得确切的查询。所以我做了一个原始查询来获取我想要的数据。有人可以帮助我如何将其转换为雄辩的laravel或查询生成器吗？

SELECT users.*,chat.*
FROM users
LEFT JOIN
  (SELECT a.customer_id,a.time,b.content
   FROM
     (SELECT customer_id,MAX(datetimestamp) TIME
      FROM chat_messages
      GROUP BY customer_id) a
   JOIN chat_messages b ON a.customer_id = b.customer_id
   AND a.time = b.datetimestamp) chat ON users.id = chat.customer_id
WHERE users.customer_role != 0
ORDER BY TIME DESC

解决方法

我认为您正在尝试为每个用户获取最新的聊天消息，可以使用左联接重写查询以选择每个组的最新记录，这样更容易以laravel的查询生成器格式转换此类查询

SQL

select u.*,c.*
from users u
join chat_messages c on u.id = c.customer_id
left join chat_messages c1 on c.customer_id = c1.customer_id and c.datetimestamp < c1.datetimestamp
where c1.customer_id is null
    and u.customer_role != 0
order by c.datetimestamp desc

查询生成器

DB::table('users as u')
  ->select('u.*,c.*')
  ->join('chat_messages as c','u.id','=','c.customer_id' )
  ->leftJoin('chat_messages as c1',function ($join) {
        $join->on('c.customer_id','c1.customer_id')
             ->whereRaw(DB::raw('c.datetimestamp < c1.datetimestamp'));
   })
  ->whereNull('c1.customer_id')
  ->where('u.customer_role','!=',0)
  ->orderBy('c.datetimestamp','desc')
  ->get();

Reference:Laravel Eloquent select all rows with max created_at