Question

我下面有两个相似的表

如果我们的购物车表如下，

id     date     title
---------------------
1      2016       a
2      2015       b
3      2018       c
4      2019       dd

产品表如下所示

id     cart_id    status
--------------------------
1         2        paid
2         2        paid
3         1        paid
4         1        unpaid
5         1        paid
6         3        paid
7         3        paid

预期输出（仅支付了所有产品的购物车）为

cart_id    status
-------------------
   3        paid
   2        paid

Answer 1

尝试使用此调整后的代码-

SELECT carts.id,'Paid' Status
FROM carts
INNER JOIN product on carts.id = product.cart_id
GROUP BY carts.id
HAVING COUNT(Status) = SUM(CASE WHEN status = 'Paid' THEN 1 ELSE 0 END)

Answer 2

这似乎是最简单的逻辑：

select p.cart_id, p.status
from product p
where p.status = 'paid';

Answer 3

您可以group by cart_id并使用HAVING子句，条件是'unpaid'列中没有status值：

select cart_id, max(status) status
from product
group by cart_id
having sum(status = 'unpaid') = 0

我假设在列status中唯一可能的值是'paid'和'unpaid'，因此您可以使用max(status) status作为第二个值，而不是'paid' status专栏。
或不存在：

select distinct p.cart_id, p.status
from product p
where not exists (
  select 1 from product
  where cart_id = p.cart_id and status = 'unpaid'
)

请参见demo。
结果：

| cart_id | status |
| ------- | ------ |
| 2       | paid   |
| 3       | paid   |

MySQL：如何选择已支付所有产品的所有购物车

3 个答案: