我的應用程序返回錯誤，我需要知道原因。我提出了一個 AJAX 請求，以獲取他在熱賣或特價商品中選擇的所有產品，并且...如果我刪除此查詢功能，它會返回數據。如何將此請求選中框傳遞給此函數？函數 App\Http\Controllers\Website\AllProductController::App\Http\Controllers\Website\{closure}() 的參數太少，1 傳入 /...../vendor/laravel/framework/src/Illuminate /Database/Eloquent/Builder.php 在第 226 行，預計正好 2public function ajax_category(Request $request)    {        if(isset($request->price) && isset($request->categories_id))        {        $product_category = $request->categories_id;        // change the value from string to array.        if (isset($request->selectedbox) && $request->selectedbox !='') {            $pairs = $request->selectedbox;            $newArray = explode(",", $pairs);        }        if (!empty($request->categories_id)) {            $max = $request->max;            $min = $request->min;        } else {            $min = product_model::where('pactive', 1)->select('MIN("price")')->first();            $max = product_model::where('pactive', 1)->select('MAx("price")')->first();        }        // change the value from string to array.        if (!empty($request->priceRange)) {            $currentRange = $request->priceRange;            $priceArray = explode(",", $currentRange);            $firstPrice = $priceArray[0];            $secondPrice = $priceArray[1];        } else {            $firstPrice = $min;            $secondPrice = $max;        }        $products = product_model::where('category', $request->categories_id)            ->whereBetween('price', [$firstPrice, $secondPrice]);        if (isset($request->selectedbox) && $request->selectedbox !='') {            $products = $products->where(function ($query,Request $request) {                $pairs = $request->selectedbox;                $newArray = explode(",", $pairs);                $query->whereIn('poffertype',implode(',', $newArray))                    ->orwhereIn('brand', implode(',', $newArray))                    ->orwhereIn('brand_ar', implode(',', $newArray));            });        }