Question

我有这种雄辩：

$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;

但是它返回了我的错误：General error: 1111 Invalid use of group function (SQL: select * from transactions where 0 = 1

如何让所有上次交易的用户等于过去30天？

以下是您需要的原始查询：

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。

由于

Answer 1

您可以在不必使用DB::raw()查询的情况下执行此操作。

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;

Answer 2

现在我尝试的是我按递减顺序对表进行排序，以便在按用户ID分组时评估最新记录...

$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;

Answer 3

无需进行原始查询。

$users = [1,2,3,4,5];
$transactions = Transaction::whereIn('user_id', $users)
                ->whereDate('created_at', '=', Carbon::today()->subDays(30)->toDateString())->get();

return $transactions;

Answer 4

你必须使用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;

Laravel雄辩，其中最大日期等于过去30天

4 个答案: