Question

我一直在研究几个表格，通过一个相当复杂的关系（我正在尝试清理，但我仍然需要通过我的Laravel从数据中做出报告）。

目前，我可以使用以下SQL查询将数据提取到我的MySQL数据库：

SELECT
customers.id,
customers.customer_name,
SUM(shipments.balance) AS shipmentBalance

FROM customers

LEFT JOIN shipments 
ON customers.id = shipments.bill_to  
AND balance > (SELECT IFNULL(SUM(payments_distributions.amount),0)                
    FROM payments_distributions               
    WHERE payments_distributions.shipment_id = pro_number)
GROUP BY customers.id, customers.customer_name
ORDER BY shipmentBalance DESC
LIMIT 5;

我只是不确定如何将它正确地重写到Laravel Eloquent所需的whereRaw或DB :: raw语句中，因为我以前的尝试都失败了。

更新

这是我尝试过的最接近的解决方案：

DB::table('customers')
        ->select('customers', DB::raw('SUM(shipments.balance) AS shipmentBalance'))
        ->leftJoin(
                 DB::raw('
                        (select shipments 
                        ON customers.id = shipments.bill_to
                        AND balance > (SELECT IFNULL(SUM(payments_distributions.amount),0) 
                            FROM payments_distributions 
                            WHERE payments_distributions.shipment_id = pro_number)'))
        ->groupBy('customers.id')
        ->orderByRaw('shipmentBalance DESC')
        ->limit(5)
        ->get();

更新2

编辑Dom：

根据您的回答使用所有内容，我得到以下回复：

SQLSTATE[42S22]: Column not found: 1054 Unknown column '' in 'on clause' (SQL: select customers.id, customers.customer_name,SUM(s.balance) AS shipmentBalance from `customers` left join `shipments` as `s` on `customers`.`id` = `s`.`bill_to` and s.balance > (SELECT IFNULL(SUM(payments_distributions.amount),0) FROM payments_distributions WHERE payments_distributions.shipment_id = s.pro_number) = `` group by `customers`.`id`, `customers`.`customer_name` order by SUM(s.balance) DESC limit 5)

但是如果我删除这一部分，它会显示页面和客户（虽然顺序错误，因为我删除了一个必要的组件：

$join->on(DB::raw('s.balance > 
            (SELECT IFNULL(SUM(payments_distributions.amount),0)                
            FROM payments_distributions               
            WHERE payments_distributions.shipment_id = s.pro_number)
                                            '));

有什么我可以提供给你的具体陈述来处理你的整个答案吗？

Answer 1

如果没有包含关系的模型或能够在此特定项目上进行测试，这是我能够考虑执行任务的最有说服力的方式。

从客户模型开始的好处是，您将拥有laravel collection，并且可以根据需要paginate。另请查看eloquent docs，它们可以帮助您了解所有不同的选项。希望他的帮助。

P.S。首先在控制器中使用您的模型，或者使用以下任何位置放置此查询：

sys/dirent.h

查询

use App\Customer

Answer 2

使用此：

DB::table('customers')
    ->select('customers.id', 'customers.customer_name', DB::raw('SUM(shipments.balance) AS shipmentBalance'))
    ->leftJoin('shipments', function($join) {
        $join->on('customers.id', 'shipments.bill_to')
            ->where('balance', '>', function($query) {
                $query->selectRaw('IFNULL(SUM(payments_distributions.amount),0)')
                    ->from('payments_distributions')
                    ->where('payments_distributions.shipment_id', DB::raw('pro_number'));
            });
    })
    ->groupBy('customers.id', 'customers.customer_name')
    ->orderByDesc('shipmentBalance')
    ->limit(5)
    ->get();

将SQL转换为Laravel Eloquent语句

更新

更新2

2 个答案: