函数的渐近分析(Asymptotic analysis of functions)

 您应该插入oc_product_option表，而不是加入它。 这只会返回已有选项的产品，而不是该类别中的所有产品。 
 
INSERT INTO oc_product_option (product_id, option_id, value, required)
SELECT p.product_id, 15 AS option_id, '' AS value, 0 AS required
FROM oc_product AS p
JOIN oc_product_to_category AS c ON o.product_id = c.product_id
WHERE c.category_id = 98
 
 实际上，甚至没有任何理由加入oc_product ，因为你没有使用该表中的任何内容。 你可以这样做： 
 
INSERT INTO oc_product_option (product_id, option_id, value, required)
SELECT product_id, 15 AS option_id, '' AS value, 0 AS required
FROM oc_product_to_category
WHERE category_id = 98

You should be inserting into the oc_product_option table, not joining with it. That will only return products that already have options, not all the products in the category.
 
INSERT INTO oc_product_option (product_id, option_id, value, required)
SELECT p.product_id, 15 AS option_id, '' AS value, 0 AS required
FROM oc_product AS p
JOIN oc_product_to_category AS c ON o.product_id = c.product_id
WHERE c.category_id = 98
 
Actually, there isn't even any reason to join with oc_product, since you're not using anything from that table. You can just do:
 
INSERT INTO oc_product_option (product_id, option_id, value, required)
SELECT product_id, 15 AS option_id, '' AS value, 0 AS required
FROM oc_product_to_category
WHERE category_id = 98