从订单中选择最后一个订单状态

Question

我有桌子：

命令：

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                 ...

如何获得所有order_id，其中该订单的最后id_order_state（最后，我的意思是说MAX（id_history）或MAX（date_add））不等于1或3？

Answer 1

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

Answer 2

另一个可能的解决方案：

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

Answer 3

我认为他所追求的是完成的订单……即订单的最终状态，而不是那些专门排除1和3的订单。 无论状态码如何，第一个预查询都应为最大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

只是为了重新显示您的数据...我已按ORDER标记了最后一个SEQUENCE（ID_History）...这就是PREQUERY将要返回的内容...

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”将包含以下子集

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，将其结果重新结合回到这两个元素上的订购历史记录中……却添加了排除“订购状态”的1,3条目的条件。

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，所有加入订单的结果都将具有一个ID，并且仅与Order_ID上的orders表匹配就可以得到所需的结果。

Answer 4

我之所以使用“我的问题的答案”，是因为我需要发布您查询的结果。 所以。

不幸的是，并非您的所有回答都能奏效。 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，目前还没有火箭科学：

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

现在为简单起见，以免更改查询以获取Last State不等于3的 Orders。

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查询：

通过RedFilter作出查询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

所以那不是真的

Tom H.提出的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

所以那不是真的

任何建议表示赞赏。

编辑

感谢Andriy M.评论，我们有适当的解决方案 。 它是对Tom H的修改。query的所有外观如下：

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：

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

所以这是真的