大型菜单的图像

这个想法是，在每个主要类别的菜单中，如果有产品，则显示其中包含的第一个产品的图像。如果产品存在，则显示，但如果在下一次迭代中，产品的下一个主要类别不存在，则显示上一个的图像，并且应该有一个存根。告诉我，有什么问题？

//определение категорий
            $sql = "SELECT name,id FROM category WHERE parrent =''";
            if($result = $conn->query($sql)){ 
                foreach($result as $row){ 

                    echo "<div class=\"uk-card uk-flex uk-flex-center uk-flex-middle\">";
                    echo "<div class=\"menu_items\"><div>";

                    $menu_category_name = $row["name"]; 
                    $menu_category_id = $row["id"];

                        //определение картинки главной категории
                        $sql = "SELECT id FROM products WHERE category='$menu_category_id' LIMIT 1";
                        if($result = $conn->query($sql)){
                            foreach($result as $row){

                            $firstproduct_id = $row["id"];

                            if ($firstproduct_id != '') { 

                                $sql = "SELECT image FROM images WHERE product_id ='$firstproduct_id' LIMIT 1";
                                if($result = $conn->query($sql)){ 
                                foreach($result as $row){ 

                                    $categoryimage = $row["image"];

                                }} else { echo "Ошибка: " . $conn->error;}

                            } else {

                                    $categoryimage = "../ui/img/noimage.jpg";
                                    
                            }
    

                            }} else { echo "Ошибка: " . $conn->error;} 

       
                    echo "<ul class=\"uk-nav-default\" uk-nav style=\"padding-left:14px;\"><li><a href=\"category.php?id=$menu_category_id\"><b>$menu_category_name</b></a></li></ul>";

                    //определение максимального количества субкатегорий
                    $sql = "SELECT name,id FROM category WHERE parrent ='$menu_category_id'";
                    if($result = $conn->query($sql)){
                    $rowsCount = $result->num_rows; // количество полученных строк
                    }

                    //определение субкатегорий
                    $sql = "SELECT name,id FROM category WHERE parrent ='$menu_category_id' LIMIT 4";
                    if($result = $conn->query($sql)){ 
                    foreach($result as $row){  

                        $menu_category_name_sub = $row["name"]; 
                        $menu_category_id_sub = $row["id"];

                        echo "<ul class=\"uk-nav-sub\"><li><a href=\"category.php?id=$menu_category_id_sub\"> - $menu_category_name_sub</a></li></ul>";
                        
                    } } else { echo "Ошибка: " . $conn->error;}


                    if ($rowsCount>=5){
                        echo "<ul class=\"uk-nav-sub\"><li><a href=\"category.php?id=$menu_category_id\"> - Смотреть все</a></li></ul>";
                    } 


                    echo "</div><div><img style=\"max-width:150px; object-fit: cover;\" src=\"$categoryimage\"></div></div>";
                    echo "</div>";
     
            } } else { echo "Ошибка: " . $conn->error;}