Php 数据库查询生成器使用关系

我有一对多的关系。组织与预订。在预订表外键组织id.组织表外键用户id.用户表我有列名|公司|账户|电话|号

组织表 控制器查询 我想要的是

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在对象内部而不是组织\u id我需要oranization.user.name、organization.user.account。。。使用关系？？如何实现这一点

您可以尝试加入这3个表 试一试

良好的实践是，应该将表名放在每一列之前，比如user.name而不是name only。

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希望能有帮助

使用适当的联接条件将

组织

表联接到查询中<代码>$bookings=DB:：table（'bookings'）->join（'organization'，'organization.id'，'='，'bookings.organization_id'）->选择（…。别忘了更改分组表达式。bookings->organizations->users，你能编辑我所有的代码吗！！

id|user_id|....

$bookings = DB::table('bookings')
                    ->select([
                        'organization_id', 
                        DB::raw('CAST(created_at AS DATE) as day'),
                        DB::raw("SUM(CASE WHEN payment_gateway = 'LocalBook' THEN 1 ELSE 0 END) AS bookings_localbook_per_day"),
                        DB::raw("SUM(CASE WHEN payment_gateway = 'LocalBook' THEN commision_i_get ELSE 0 END) AS commision_localbook_per_day"),
                        DB::raw("SUM(CASE WHEN payment_gateway = 'selfBook' THEN 1 ELSE 0 END) AS bookings_selfbook_per_day"),
                        DB::raw("SUM(CASE WHEN payment_gateway = 'selfBook' THEN commision_i_get ELSE 0 END) AS commision_selfbook_per_day"),

                        DB::raw("SUM(CASE WHEN payment_gateway = 'paypal' THEN 1 ELSE 0 END) AS bookings_international_per_day"),
                        DB::raw("SUM(CASE WHEN payment_gateway = 'paypal' THEN commision_i_get ELSE 0 END) AS commision_international_per_day"),
                        DB::raw('SUM(commision_i_get) AS total_comission_per_day'),
                    ])
                    ->where('pay', '>', 0)
                    ->groupBy(['organization_id', DB::raw('CAST(created_at AS DATE)')])
                    ->get()
                    ->toArray();

DB::table('bookings')
    ->join('organizations','organizations.id','=','bookings.organization_id')
    ->join('user','user.id','=','organizations.user_id')
    ->select([
                       'user.account',
                       'user.name',
                        ....])