Question

我有三张桌子 cards (id,...) order_cards (card_id, order_id....) transactions (order_card_id, status, ....)

我想让所有最近7次交易失败的卡

我正在尝试类似的事情

select  cards.*,
(
select sum(case tmp1.status when 'fail' then 1 else 0 end) from  
(select transactions.status from order_cards
left join transactions on transactions.order_card_id = order_cards.id
where order_cards.card_id = cards.id
order by transactions.id desc limit 7
) as tmp1
) as total_fail
from cards

group by cards.id
having total_fail > 5

遇到此错误

 Unknown column 'cards.id' in 'where clause'

具有上述查询的

问题是父级ID在第2级子查询中不起作用。尝试使用具有count的hading子句，但不能与limit一起使用任何建议，谢谢

已经尝试过

Use column of parent query in subquery of a subquery

MYSQL - Get all records that have more than 1 record for the same id

Answer 1

这应该有效。 Select c.*, SUM(t.status = 'FAILED') AS failed_txns from cards c, order_cards oc, transactions t where t.status='FAILED' and oc.order_id = t.order_card_id and c.id = oc.card_id group by c.id having failed_txns >= 7

Answer 2

我认为parent-id不能在FROM子句的派生表中使用，但仍可以在WHERE子句和子子查询的SELECT列表中访问。因此，在您的情况下，您可以找到card_id的第8个最近的transaction.id（LIMIT 7,1），然后根据该transaction.id找到SUM() ：（请注意，同一事务中少于8个事务card_id，LIMIT 7,1将返回NULL，在这种情况下，所有交易都应计入，因此如下：t1.id > IFNULL((...), 0)）：

SELECT  c.*
,   IFNULL((
      SELECT sum(t1.status = 'fail') 
      FROM order_cards o1
      JOIN transactions t1 on t1.order_card_id = o1.id
      WHERE o1.card_id = c.id
      AND t1.id > IFNULL((
          SELECT t2.id 
          FROM order_cards o2
          JOIN transactions t2 on t2.order_card_id = o2.id
          WHERE o2.card_id = c.id
          ORDER BY t2.id DESC 
          LIMIT 7,1
      ), 0)
    ), 0) as total_fail
FROM cards c
HAVING total_fail > 5

我在自己的服务器的数据库上测试了一个类似的SQL，该SQL正常运行，并且相同的方法可能适用于您的情况。

mysql获取最近7次失败交易的所有卡记录

2 个答案: