我有一個(gè)名為“業(yè)務(wù)”的模型，一個(gè)企業(yè)可以提供多種服務(wù)。我有另一種模型，稱為“付款”。付款記錄了人們支付的服務(wù)費(fèi)用。一項(xiàng)服務(wù)可以有多個(gè)付款。我打算根據(jù)收到的付款獲取排名前 10 名和排名最差的 10 家企業(yè)。下面的代碼工作正常，但效率很低。我必須循環(huán)遍歷整個(gè)數(shù)據(jù)才能檢索我需要的信息。有更有效的方法來實(shí)現(xiàn)這一目標(biāo)嗎？$businesses = Business::with(['services'])->get();foreach($businesses as $business){    $id = $business->id;    $name = $business->display_name;    $services = $business->services;    $businessRevenue = 0;    if(count($services)>0){        foreach($services as $service){            $serviceId = $service->id;            $totalAmount = PaymentTransaction::whereHas('invoice', function($query) use ($serviceId){                $query->where('product_code_id', $serviceId);            })->where('amount_paid', ">", 0)->sum('amount_paid');            $businessRevenue= $businessRevenue + $totalAmount;        }    }    $businessArray = (object) array('id'=> $id, 'name'=> $name, 'revenue'=> $businessRevenue);        array_push($transformedBusiness, $businessArray);}$topBusiness = $bottomBusiness = $transformedBusiness;usort($bottomBusiness, function($a, $b) {return strcmp($a->revenue, $b->revenue);});usort($topBusiness, function($a, $b) {return strcmp($b->revenue, $a->revenue);});$topBusiness = array_slice($topBusiness, 0, 10);$bottomBusiness = array_slice($bottomBusiness, 0, 10);return view('report.department_performance', compact('topBusiness', 'bottomBusiness'));