Hoe kan ik voorwaardelijke aggregatie mysql converteren naar laravel query?

U moet het aggregate . opgeven functie in het hebben van clausule, kunnen we dezelfde hergebruiken die we hebben in select, zoals deze

$brand = "MAX(CASE WHEN b.attribute_code = 'brand' then b.attribute_value END)";

$model = "MAX(CASE WHEN b.attribute_code = 'model' then b.attribute_value END)";

$category = "MAX(CASE WHEN b.attribute_code = 'category' then b.attribute_value END)";

$subcategory = "MAX(CASE WHEN b.attribute_code = 'subcategory' then b.attribute_value END)";

$brandName = 'honda';

$query = Item::selectRaw("a.number, a.description, {$brand} as brand, {$model} as model, {$category} as category, {$subcategory} as subcategory")
        ->from('items as a')
        ->join('attr_maps as b','b.number','=','a.number')
        ->groupBy('a.number')
        ->havingRaw("{$brand} = ?", [$brandName])
        ->orderBy('description')
        ->paginate(10);

return $query;

BEWERKEN: Na opmerkingen

U kunt voor elke parameter als volgt uitvoeren

$query = Item::selectRaw("a.number, a.description, {$brand} as brand, {$model} as model, {$category} as category, {$subcategory} as subcategory")
    ->from('items as a')
    ->join('attr_maps as b','b.number','=','a.number')
    ->groupBy('a.number')
    ->orderBy('description');

foreach($param as $key => $value) {
     $query = $query->havingRaw("{$$key} = ?", [$value]);
}

$results = $query->paginate(10);

return $results;