我有桌子：
订单：
id_order id_customer
1 1
2 2
3 1
orders_history
id_history id_order id_order_state date_add
1 1 1 2010-01-01 00:00:00
2 1 2 2010-01-02 00:00:00
3 1 3 2010-01-03 00:00:00
4 2 2 2010-05-01 00:00:00
5 2 3 2011-05-02 00:00:00
6 3 1 2011-05-03 00:00:00
7 3 2 2011-06-01 00:00:00
order_state
id_order_state name
1 New
2 Sent
3 Rejected
4 ...
如何获取所有order_id的那个订单的最后一个id_order_state（最后我的意思是MAX（id_history）或MAX（date_add））不等于1或3？
答案 0 :(得分：4)
select oh.id_history, oh.id_order, oh.id_order_state, oh.date_add
from (
select id_order, max(date_add) as MaxDate
from orders_history
where id_order_state not in (1, 3)
group by id_order
) ohm
inner join orders_history oh on ohm.id_order = oh.id_order
and ohm.MaxDate = oh.date_add
答案 1 :(得分：1)
另一种可能的解决方案：
SELECT DISTINCT
id_order
FROM
Orders_History OH1
LEFT OUTER JOIN Orders_History OH2 ON
OH2.id_order = OH1.id_order AND
OH2.is_order_state IN (1, 3) AND
OH2.date_add >= OH1.date_add
WHERE
OH2.id_order IS NULL
答案 2 :(得分：1)
我认为他所追求的是订单是完整的......即他们的最终状态，而不是那些专门排除1和3的人。无论状态代码如何，第一个预查询都应该是最大ID
select
orders.*
from
( select oh.id_order,
max( oh.id_history ) LastID_HistoryPerOrder
from
orders_history oh
group by
oh.id_order ) PreQuery
join orders_history oh2
on PreQuery.ID_Order = oh2.id_order
AND PreQuery.LastID_HistoryPerOrder = oh2.id_history
AND NOT OH2.id_order_state IN (1, 3) <<== THIS ELIMINATES 1's & 3's from result set
join Orders <<= NOW, anything left after above ^ is joined to orders
on PreQuery.ID_Order = Orders.ID_Order
只是为了重新显示你的数据......我已经为每个ORDER标记了最后一个SEQUENCE（ID_History）......这就是PREQUERY要返回的......
id_history id_order id_order_state date_add
1 1 1 2010-01-01 00:00:00
2 1 2 2010-01-02 00:00:00
**3 1 3 2010-01-03 00:00:00
4 2 2 2010-05-01 00:00:00
**5 2 3 2011-05-02 00:00:00
6 3 1 2011-05-03 00:00:00
**7 3 2 2011-06-01 00:00:00
“PreQuery”将产生以下子集
ID_Order LastID_HistoryPerOrder (ID_History)
1 3 (state=3) THIS ONE WILL BE SKIPPED IN FINAL RESULT
2 5 (state=3) THIS ONE WILL BE SKIPPED IN FINAL RESULT
3 7 (state=2)
现在，然后将结果重新加入到这两个元素的订单历史记录中......然后添加标准以排除“订单状态”的1,3个条目。
在这种情况下，
1 would be rejected as its state = 3 (sequence #3),
2 would be rejected since its last history is state = 3 (sequence #5).
3 would be INCLUDED since its state = 2 (sequence #7)
最后，加入订单的所有内容都会产生一个ID，并且仅与Order_ID上的订单表很好地匹配，并获得所需的结果。
答案 3 :(得分：0)
我正在使用＆＃34;回答我的问题＆＃34;因为我需要发布您的查询结果。如此。
不幸的是，并非你所有的答案都有效。让我们准备测试环境：
CREATE TABLE `order_history` (
`id_order_history` int(11) NOT NULL AUTO_INCREMENT,
`id_order` int(11) NOT NULL,
`id_order_state` int(11) NOT NULL,
`date_add` datetime NOT NULL,
PRIMARY KEY (`id_order_history`)
) ENGINE=MyISAM AUTO_INCREMENT=11 DEFAULT CHARSET=latin2;
CREATE TABLE `orders` (
`id_order` int(11) NOT NULL AUTO_INCREMENT,
`id_customer` int(11) DEFAULT NULL,
PRIMARY KEY (`id_order`)
) ENGINE=MyISAM AUTO_INCREMENT=8 DEFAULT CHARSET=latin2;
INSERT INTO `order_history`
(`id_order_history`, `id_order`, `id_order_state`, `date_add`) VALUES
(1,1,1,'2011-01-01 00:00:00'),
(2,1,2,'2011-01-01 00:10:00'),
(3,1,3,'2011-01-01 00:20:00'),
(4,2,1,'2011-02-01 00:00:00'),
(5,2,2,'2011-02-01 00:25:01'),
(6,2,3,'2011-02-01 00:25:59'),
(7,3,1,'2011-03-01 00:00:01'),
(8,3,2,'2011-03-01 00:00:02'),
(9,3,3,'2011-03-01 00:01:00'),
(10,3,2,'2011-03-02 00:00:01');
COMMIT;
INSERT INTO `orders` (`id_order`, `id_customer`) VALUES
(1,1),
(2,2),
(3,3),
(4,4),
(5,5),
(6,6),
(7,7);
COMMIT;
现在，让我们为每个订单选择最后/最大状态，让我们运行简单的查询：
select id_order, max(date_add) as MaxDate
from `order_history`
group by id_order
这给了我们正确的结果，现在没有火箭科学：
id_order MaxDate
---------+-------------------
1 2011-01-01 00:20:00 //last order_state=3
2 2011-02-01 00:25:59 //last order_state=3
3 2011-03-02 00:00:01 //last order_state=2
现在为了简单起见，以免更改我们的查询以获取订单其中最后状态不等于3 。
我们期望获得id_order = 3 的一行结果。
因此，让我们测试一下我们的问题：
由 RedFilter 制作的QUERY 1：
select oh.id_order, oh.id_order_state, oh.date_add
from (
select id_order, max(date_add) as MaxDate
from `order_history`
where id_order_state not in (3)
group by id_order
) ohm
inner join `order_history` oh on ohm.id_order = oh.id_order
and ohm.MaxDate = oh.date_add
结果：
id_order id_order_state date_add
-------------------------------------------------
1 2 2011-01-01 00:10:00
2 2 2011-02-01 00:25:01
3 2 2011-03-02 00:00:01
所以不正确
由Tom H撰写的QUERY 2。：
SELECT DISTINCT OH1.id_order
FROM order_history OH1
LEFT OUTER JOIN order_history OH2 ON
OH2.id_order = OH1.id_order AND
OH2.id_order_state NOT IN (3) AND
OH2.`id_order_history` >= OH1.`id_order_history`
WHERE
OH2.id_order IS NULL
结果：
id_order
--------
1
2
所以不正确
任何建议表示赞赏。
修改
感谢Andriy M.的评论，我们有正确的解决方案。它是Tom H.查询的修改，应该如下所示：
SELECT DISTINCT
OH1.id_order
FROM
order_history OH1
LEFT OUTER JOIN order_history OH2 ON
OH2.id_order = OH1.id_order
AND OH2.date_add > OH1.date_add
WHERE OH1.id_order_state NOT IN (3) AND OH2.id_order IS NULL
编辑2：
由 DRapp 制作的QUERY 3：
select
distinct orders.`id_order`
from
( select oh.id_order,
max( oh.id_order_history ) LastID_HistoryPerOrder
from
order_history oh
group by
oh.id_order ) PreQuery
join order_history oh2
on PreQuery.id_order = oh2.id_order
AND PreQuery.LastID_HistoryPerOrder = oh2.id_order_history
AND NOT oh2.id_order_state IN (1,3)
join orders
on PreQuery.id_order = orders.id_order
结果：
id_order
--------
3
所以它终于真实