Question

我是laravel 5 sql查询构建的新手，我面临将sql转换为laravel 5.x sql查询的挑战。

以下是正常的sql：

select users.name as name, count(ticket.id) as total, sum(ticket.quantity * ticket.amount) as montant 
from users 
inner join program on program.driver_id = users.id
inner join line on line.id = program.line_id
inner join trip on trip.program_id = program.id
inner join ticket  on ticket.trip_id = trip.id
where program.date = $date
group by users.name

给了我需要的回复。但我的代码的laravel版本什么也没有返回。以下是我的代码

$result = \DB::table('users')
                ->select('users.name as name', \DB::raw('count(ticket.id) as total'),\DB::raw('SUM(ticket.quantity*ticket.amount) as montant'))
                ->Join('program', 'program.driver_id', '=', 'users.id')
                ->Join('line', 'line.id', '=', 'program.line_id')
                ->Join('trip', 'trip.program_id', '=', 'program.id')
                ->Join('ticket', 'trip.id', '=', 'ticket.trip_id')
                ->where('program.date', $date)
                ->groupBy('users.name')
                ->get();

提前致谢

Answer 1

您可以尝试以下查询：

use Illuminate\Support\Facades\DB;

$result = DB::table('users')
                ->selectRaw('users.name as name, count(ticket.id) as total, SUM(ticket.quantity*ticket.amount) as montant'))
                ->join('program', 'program.driver_id', '=', 'users.id')
                ->join('line', 'line.id', '=', 'program.line_id')
                ->join('trip', 'trip.program_id', '=', 'program.id')
                ->join('ticket', 'ticket.trip_id', '=', 'trip.id')
                ->where('program.date', $date)
                ->groupBy('users.name','ASC')
                ->get();

SQL查询转换为laravel查询构建

1 个答案: