我有一个雄辩的:
$users = [1,2,3,4,5];
$transactions = Transaction::whereIn('user_id', $users)
->where(DB::raw('DATE(MAX(created_at))'), Carbon::now()->subMonth()->format('Y-m-d'))->get();
return $transactions;
但它返回给我这个错误:General error: 1111 Invalid use of group function (SQL: select * from transactions where 0 = 1
我如何获得所有用户的最后一笔交易等于过去30天?
以下是原始查询以备不时之需:
select * from `transactions` where `0` = 1 and DATE(MAX(created_at)) = ?
表格结构:
Users Table: id (not autoincrement), name, gender, created_at
Transactions Table: id (not autoincrement), user_id (FK), amount, created_at
我目前的解决方案:
迭代每个用户并检查他们的最大created_at
。
感谢
您可以简单地执行此操作,而不必使用DB::raw()
查询。
试试这个:
$users = [1,2,3,4,5];
$transactions = Transaction::whereIn('user_id', $users)
->where('created_at', '=', Carbon::now()->format('Y-m-d'))
->get();
return $transactions;
不需要进行原始查询。
$users = [1,2,3,4,5];
$transactions = Transaction::whereIn('user_id', $users)
->whereDate('created_at', '=', Carbon::today()->subDays(30)->toDateString())->get();
return $transactions;
您必须使用whereBetween
$users = [1,2,3,4,5];
$transactions = Transaction::select(DB::raw('DATE(MAX(created_at))'))->whereIn('user_id', $users)
->whereBetween('created_at',array(Carbon::now(),Carbon::now()->subMonth()->format('Y-m-d')))->get();
return $transactions;
现在我尝试的是按递减顺序对表进行排序,以评估最新记录何时按用户Id分组…
$users = [1,2,3,4,5];
$transactions = Transaction::select(DB::raw('select id from transactions order by transaction.created_at desc as table1' )->leftJoin('transactions','id','=','table1.id')->whereIn('user_id', $users)
->where('created_at', '=', Carbon::now()->subMonth()->format('Y-m-d'))->groupBy('transaction.user_id')
->get();
return $transactions;