php - 通过 Woocommerce 中的变体 ID 获取属性 slug 值

Question

我正在构建自定义产品页面，但我遇到了变体问题

这是我的代码:

global $product;
$variations = $product->get_available_variations();

foreach($variations as $variation)
{

     $variation_id = $variation['variation_id'];

     $variation_obj = new WC_Product_variation($variation_id);

     //THIS IS PROBLEM FAIL TEST (It's exemple)
     $type = $variation_obj->get_attributes()->get_name();
     //-----

     $stock = $variation_obj->get_stock_quantity();
     $COD = $variation_obj->get_sku();
     $regularprice = $variation_obj->get_regular_price();
     $saleprice = $variation_obj->get_sale_price();
     $diff = $regularprice-$saleprice;

     echo "<p>Variation: " .$type."</p>";
     echo "<p>P.Code: " .$COD."</p>";
     echo "<p>Regular price: ".$regularprice."</p>";
     echo "<p>Sale price: ".$saleprice."</p>";
     echo "<p>Difference: ". $diff."</p>";
     echo "<p>Stock: ".$stock."</p>";
}

我想知道是否可以提取属性或变体的单个名称？

或者没有完整标题的变体名称？

例子:我想要“蓝色”...而不是“神奇美丽的 Pouf Lumaland - 蓝色”

Answer 1

属性 slug 值的数组包含在 $variation 数组中。所以试试这个:

global $product;
if($product->is_type('variable')){
    $variations = $product->get_available_variations();

    foreach($variations as $variation)
    {
        $variation_obj = wc_get_product($variation['variation_id']);

        // Variation can have many poduct attributes
        $attr_slugs = implode(', ', $variation['attributes']);

        $stock = $variation_obj->get_stock_quantity();
        $COD = $variation_obj->get_sku();
        $regularprice = $variation_obj->get_regular_price();
        $saleprice = $variation_obj->get_sale_price();
        $diff = $regularprice-$saleprice;

        echo "<p>Var Name: " .$attr_slugs."</p>";
        echo "<p>Var Name: " .$COD."</p>";
        echo "<p>regular price: ".$variation['display_regular_price']."</p>";

        echo "<p>sale price: ".$saleprice."</p>";
        echo "<p>risparmi: ". $diff."</p>";
        echo "<p>in stock: ".$stock."</p>";
    }
}

经过测试并有效。