所以我有两个数据库表：account, account_sub。
一个account（具有唯一ID字段account_id）可能有account_sub
account_sub包含以下字段：status (an int, can be 1-6), start_date, duration, account_id(foreign key to account)
例如：
accounts
account_id name
1 Apple
2 Orange
3 Banana
4 grape
account_subs
id account_id status date_start duration
1 1 1 1/2/09 5
2 2 5 1/2/09 10
3 1 3 1/3/09 11
4 2 6 1/3/09 13
5 1 3 1/4/09 22
6 2 6 1/4/09 19
7 3 2 1/4/09 100
8 4 6 1/4/09 40
我想找到account_sub的最新last_update_date列表（由account决定）所有account_sub没有status = 1-4的{{1}}
这就是假设的结果：
account_subs:
account_id account_sub_id
2 6
4 8
因为只有帐户2和4没有任何状态为1-4的account_sub，而帐户子6的开始日期+持续时间是最大的，最新的，account_sub 8也是帐户4的最新帐号。 /强>
因此，account与account_sub status != 1-4之间的account_sub应该在结果中包含最新的select account_sub
from account_sub, account
where account_sub.account_id = account.account_id
and account_sub.status != 1
and account_sub.status != 2
and account_sub.status != 3
and account_sub.status != 4
。
这就是我现在所拥有的：
where我认为这是不正确的，它不能得到我想要的东西，我想我需要account_sub中更复杂的条款，我不知道如何计算最近的With a as (
SELECT name_id
, take
, id_user
FROM ((Proizv
INNER JOIN Izdan
ON Proizv.id = Izdan.id_proizv)
INNER JOIN Exzemp
ON Izdan.id = Exzemp.id_izdan)
INNER JOIN log
ON Exzemp.id = log.id_exzemp
WHERE take IN ( SELECT TOP 2 take
FROM log AS s
WHERE s.id_user = log.id_user
ORDER BY take DESC )
),
b as (
SELECT COUNT(name_id) AS cname
, name_id
FROM ((Proizv
INNER JOIN Izdan
ON Proizv.id = Izdan.id_proizv)
INNER JOIN Exzemp
ON Izdan.id = Exzemp.id_izdan)
INNER JOIN log
ON Exzemp.id = log.id_exzemp
GROUP BY name_id
)
select a.*, b.cname
from a inner join b
on a.name_id = b.name_id
。
答案 0 :(得分：1)
select account_id, id
from
(select account_sub.*, ( select count(*) from account_sub sub
where date_add(sub.date_start, interval sub.duration day) >=
date_add(account_sub.date_start, interval account_sub.duration day)
and sub.account_id = account_sub.account_id) as seq
from
account
inner join
account_sub
on account.account_id = account_sub.account_id
where
account.account_id not in (select distinct accsub.account_id
from account_sub accsub
where
accsub.status in (1, 2, 3, 4))) t
where seq = 1 order by account_id;