Laravel 子查询答案

【问题标题】：Laravel subqueryLaravel 子查询
【发布时间】：2016-04-07 05:52:38
【问题描述】：

我想在 laravel 5.2 中编写这个查询

SELECT b.id,
       TotalP,
       b.booking_amount
FROM booking b
LEFT JOIN
  (SELECT sum(amount) AS TotalP,
          booking_id
   FROM payment
   GROUP BY booking_id) AS T ON b.id = T.booking_id
WHERE COALESCE(TotalP, 0) < b.booking_amount

我的问题与这篇文章有关。我在搜索和研究后写了一个查询，但它不起作用，需要更多约束

 $result = DB::table('my_booking')
        ->select('booking_name')
        ->leftJoin(DB::raw('(SELECT booking_id,sum(amount) as TotalP FROM `my_payment` GROUP BY booking_id) TotalPayment'), function($join)
        {
            $join->on('my_booking.id', '=', 'TotalPayment.booking_id');
        })
        ->get();

Sql query to get data diffrence of total in 2 tables

【问题讨论】：

标签： php mysql database laravel

【解决方案1】：

你可以试试这个，

$booking_payments = Booking::with('Payment')->find(1);

$total = 0;
foreach($booking_payments->payment as $booking_payment){
$total += $booking_payment->amount;
}

if($booking_payments->booking_amount == $total){
// if the total and booking_amount is equal
}

【讨论】：

【解决方案2】：

这应该可以在 Laravel 中运行，并为您提供与 MySQL 查询相同的结果。我将 COALESCE 移到子查询选择区域中，这样您就不必在 Laravel 中编写原始 DB where 语句。

$sql_subquery = "(SELECT COALESCE(SUM(amount),0) AS TotalP,
                          booking_id
                   FROM payment
                   GROUP BY booking_id) AS T";


$result = DB::table('booking AS b')
            ->leftJoin(DB::raw($sql_subquery), 'b.id', '=', 'T.booking_id')
            ->where('T.TotalP','<', 'b.booking_amount')
            ->select('b.id','T.TotalP','b.booking_amount')
            ->get();

【讨论】：