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试过了，但我不能正常工作。

$sql = mysqli_query($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()

如何将原始查询 MYSQL 转换为 Laravel 框架

How to convert raw query MYSQL to Laravel framework

php

mysql

laravel

laravel-5.8