Question

我有以下选择

$repairs = DB::select(DB::raw('
            select r.id repair_id, w.name, w.surname
                from repairs r
                    join repair_worker rw on r.id = rw.repair_id
                    join workers w on w.id = rw.worker_id
                where r.vehicle_id = ?
                group by w.name, w.surname, r.id
           '),[$vehicle->id]);

我试图将w.name和w.surname放入一个字段中，例如用逗号分隔。有一个函数string_agg(column, ',')但我没有运气让它工作。我总是得到语法错误。

我的尝试：

$repairs = DB::select(DB::raw('
                select r.id repair_id, string_agg(w.name, w.surname, ', ') workers
                    from repairs r
                        join repair_worker rw on r.id = rw.repair_id
                        join workers w on w.id = rw.worker_id
                    where r.vehicle_id = ?
                    group by r.id
          '),[$vehicle->id]);

你会怎么做？

我正在使用最新的Postgresql 9 +

Answer 1

使用||运算符执行字符串连接，然后汇总此结果。

$repairs = DB::select(DB::raw("
    select r.id repair_id, string_agg(w.name || ' ' || w.surname, ', ') workers
    from repairs r
        join repair_worker rw on r.id = rw.repair_id
        join workers w on w.id = rw.worker_id
    where r.vehicle_id = ?
    group by r.id
    "),[$vehicle->id]);

Answer 2

试试这个吗？

$repairs = DB::select(DB::raw("
                select r.id repair_id, string_agg(concat(w.name, ',',w.surname), ', ') workers
                    from repairs r
                        join repair_worker rw on r.id = rw.repair_id
                        join workers w on w.id = rw.worker_id
                    where r.vehicle_id = ?
                    group by r.id
          "),[$vehicle->id]);

String_agg用于多列

2 个答案: