Question

我要通过以下方式从严格表中检索数据：

$data = Accommodation::join("hotels", "accommodations.hotel_id", "=", "hotels.id")
        ->join("room_types", "accommodations.room_type", "=", "room_types.id")
        ->join("country", "accommodations.country_id", "=", "country.id")
        ->join("accommodation_accreditation", "accommodations.id", "=", "accommodation_accreditation.accommodation_id")
        ->join("accreditation", 'accreditation.id', "=", "accommodation_accreditation.accreditation_id")
        ->select('hotels.name as Hotel' ,'room_types.type as Room', 'price_per_night as Price per Night', 'check_in as Check In', 'check_out as Check Out', 'nights as Nights', 'sub_total as Sub Total', 'country.name as Country', 'accreditation.full_name as Occupants')
        ->get()->toArray();

我会收到一个房间里每个人的一行。

          Room           Price per Night    Check In    Check Out    Nights   Sub Total   Country       Occupants      
 ---------------------- ----------------- ------------ ------------ -------- ----------- --------- ------------------- 
  Twin/Double standart               160   2019-04-10   2019-04-15        5         800   UKR       Vlad Timochenko  
  Twin/Double standart               160   2019-04-10   2019-04-15        5         800   UKR       Selena Viachenko

我需要隐藏住户的名字，并像这样将它们排成一排：

          Room           Price per Night    Check In    Check Out    Nights   Sub Total   Country               Occupants               
 ---------------------- ----------------- ------------ ------------ -------- ----------- --------- ------------------------------------ 
  Twin/Double standart               160   2019-04-10   2019-04-15        5         800   UKR       Vlada Nikolchenko, Olena Diachenko

这是我正在讨论的表的模式

希望很清楚。如果您能帮上忙，那就太好了。

Answer 1

说实话，我从不同的地方找到了完整的解决方案。

当萨姆感到很轻松时，我使用了GROUP_CONCAT。我的代码如下：

$data = Accommodation::join("hotels", "accommodations.hotel_id", "=", "hotels.id")
        ->join("room_types", "accommodations.room_type", "=", "room_types.id")
        ->join("country", "accommodations.country_id", "=", "country.id")
        ->join("accommodation_accreditation", "accommodations.id", "=", "accommodation_accreditation.accommodation_id")
        ->join("accreditation", 'accreditation.id', "=", "accommodation_accreditation.accreditation_id")
        ->select('hotels.name as Hotel' ,'room_types.type as Room', 'price_per_night as Price per Night', 'check_in as Check In', 'check_out as Check Out', 'nights as Nights', 'sub_total as Sub Total', 'country.name as Country', DB:raw("GROUP_CONCAT('accreditation.full_name SEPRATOR ',') as `Occupants`"))
        ->groupBy('accommodation.id')
        ->get()->toArray();

我要警惕的另一件事是在Occupants标签上使用反引号，否则SEPARATOR无法正常工作。

还有其他类似的线程，所以这是重复的。

Group_concat - laravel eloquent how to use GROUP_CONCAT in laravel

用laravel雄辩地合并/合并数据行

1 个答案: