在 laravel 框架中使用 whereraw/where 中的 SUM

Question

我有名为 PlanCost Discount AmountPaid 的列和一个包含 Due

的标签

在这里，我试图只显示 PlanCost != AmountPaid

这是我的全部查询

 $getID = isset($inputArray['id']) ? $inputArray['id'] : 0;
    $query = TelePlanSelect::join('tele_plan', 'tele_plan.id', '=', 'tele_plan_select.teleplan_id'  )
        ->leftjoin('tele_payment_defs', 'tele_payment_defs.telereg_id', '=', 'tele_plan_select.telereg_id')
        ->leftjoin('tele_payment_items', function ($join){
            $join->on('tele_payment_items.telepaymentdefs_id', '=', 'tele_payment_defs.id')
                ->on('tele_payment_items.teleplan_id', '=' ,'tele_plan.id');
        } )

        ->selectRaw('tele_plan.id as `PlanID`,' .
            'tele_plan.plan_name as `PlanName`,' .
            'tele_plan.plan_cost as `PlanCost`,' .
            'tele_plan.plan_details as `PlanDetails`,'.
            'sum(tele_payment_items.amount) as  `AmountPaid` ,'.
            'COALESCE(sum(tele_payment_items.discount),0) as `Discount`,'.
            '(COALESCE(tele_plan.plan_cost,0) -  sum(COALESCE(tele_payment_items.discount,0))) - sum(coalesce(tele_payment_items.amount, 0)) as  `Due` '

        )
        ->where('tele_plan_select.telereg_id', $getID)
        ->groupBy('tele_plan.id')
      ->where(' tele_plan.plan_cost', '!=', 'sum(tele_payment_items.amount)');

我不确定我们是否可以在 where/whereraw 中使用 sum 有人可以帮助我吗？

编辑 1： {"message":"SQLSTATE[HY000]: General error: 1111 Invalid use of group function (SQL: select count(*) as aggregate from (select tele_plan.id asPlanID,tele_plan.plan_name asPlanName,tele_plan.plan_cost asPlanCost,tele_plan.plan_details asPlanDetails,sum(tele_payment_items.amount) asAmountPaid,COALESCE(sum(tele_payment_items.discount),0) asDiscount,(COALESCE(tele_plan.plan_cost,0) - sum(COALESCE(tele_payment_items.discount,0))) - sum(coalesce(tele_payment_items.amount, 0)) asDuefromtele_plan_selectinner jointele_planontele_plan.id=tele_plan_select.teleplan_idleft jointele_payment_defsontele_payment_defs.telereg_id=tele_plan_select.telereg_idleft jointele_payment_itemsontele_payment_items.telepaymentdefs_id=tele_payment_defs.idandtele_payment_items.teleplan_id=tele_plan.idwheretele_plan_select.telereg_id = 8 and tele_plan.plan_cost <> sum(tele_payment_items.amount) group bytele_plan.id) as a) @ C:\xampp\htdocs\halframework\vendor\laravel\framework\src\Illuminate\Database\Connection.php:625","status":"failed"}

Answer 1

您可以使用像 sum 这样的原始 sql 函数，但您必须将它们包装在 laravel 的 \DB::raw() 函数中，这样它们就不会被转义。

对于您的 Where 子句，您应该执行以下操作

->where('tele_plan.plan_cost', '<>', \DB::raw('sum(tele_payment_items.amount)'));

您也可以使用laravel的WhereRaw函数来编写您自己的条件：

->whereRaw('tele_plan.plan_cost <> sum(tele_payment_items.amount)')

更新： 该错误表明 GroupBy 存在问题。之所以这样是因为a sum 只能在group by 之后做，而a where 不行。因此，必须将其添加到 HAVING clause.

中

将 where 替换为 having，这应该有效：

->havingRaw('tele_plan.plan_cost <> sum(tele_payment_items.amount)');

在 laravel 框架中使用 whereraw/where 中的 SUM

using SUM in whereraw/where in laravel framework

php

mysql

laravel