下面的代碼使用 2 個表“category_path”和“category_description”來獲取所有類別和子類別的 id=>name。我不擅長 mySql，所以如果你能幫助我，我將不勝感激。在此函數中，我還需要從名為“category”的其他（第三個）表中獲取“cat_name”列的值https://i.stack.imgur.com/1PC4A.jpgpublic function getCategories($data = array()) {    $sql = "SELECT cp.category_id AS category_id,                 GROUP_CONCAT(cd1.name ORDER BY cp.level SEPARATOR '  >  ') AS name,                 c1.parent_id, c1.sort_order             FROM " . DB_PREFIX . "category_path cp                 LEFT JOIN " . DB_PREFIX . "category c1 ON (cp.category_id = c1.category_id)                 LEFT JOIN " . DB_PREFIX . "category c2 ON (cp.path_id = c2.category_id)                 LEFT JOIN " . DB_PREFIX . "category_description cd1 ON (cp.path_id = cd1.category_id)                 LEFT JOIN " . DB_PREFIX . "category_description cd2 ON (cp.category_id = cd2.category_id)              WHERE cd1.language_id = '" . (int)$this->config->get('config_language_id') . "'             AND cd2.language_id = '" . (int)$this->config->get('config_language_id') . "'";