如何将原始查询 MYSQL 转换为 Laravel 框架

Question

我不知道如何在 Laravel 5.8 中使用下面的代码，我已经尝试过，但无法正常工作。$sql = mysqli_query($db, "SELECT *, SUM(amount) AS SumBudget FROM messages GROUP BY contact_phone ORDER BY SUM(amount) DESC LIMIT 3");$orderList = 0;while ($data = mysqli_fetch_assoc($sql)){    $orderList++;    $user = $data['user'];     $cost = $data['SumBudget'];    if ($orderList == 1) {        printf("%d. %s %d‎฿<br>", $orderList, $user, $cost);    }    else if ($orderList == 2)    {        printf("%d. %s %d‎฿<br>", $orderList, $user, $cost);    }    else     {        printf("%d. %s %d‎฿<br>", $orderList, $user, $cost);    }}我的代码已经试过了。$data = DB::table('messages')        ->where('phone_number', $request->phone_number)        ->select(DB::raw('SUM(amount) as cost'))        ->groupBy(DB::raw('contact_phone'))        ->orderBy(DB::raw('SUM(amount)', 'DESC'))        ->limit(3)        ->get();谁能帮帮我，谢谢。

Answer 1

我认为它可以帮助你：

DB::table('messages')
    ->select('column_name', /* ... */, DB::raw('SUM(amount) AS cost'))
    ->where('phone_number', $request->phone_number) // if you want to filter
    ->groupBy('contact_phone')
    ->orderBy('cost', 'DESC')
    ->take(3)
    ->get()