從訂單中選擇最后一個訂單狀態

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                 ...

如何獲得所有order_id，其中該訂單的最后id_order_state（最后，我的意思是說MAX（id_history）或MAX（date_add））不等於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

另一個可能的解決方案：

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

我認為他所追求的是完成的訂單……即訂單的最終狀態，而不是那些專門排除1和3的訂單。 無論狀態碼如何，第一個預查詢都應為最大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

只是為了重新顯示您的數據...我已按ORDER標記了最后一個SEQUENCE（ID_History）...這就是PREQUERY將要返回的內容...

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”將包含以下子集

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)

現在，將其結果重新結合回到這兩個元素上的訂購歷史記錄中……卻添加了排除“訂購狀態”的1,3條目的條件。

在這種情況下，

   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，所有加入訂單的結果都將具有一個ID，並且僅與Order_ID上的orders表匹配就可以得到所需的結果。

Answer 4

我之所以使用“我的問題的答案”，是因為我需要發布您查詢的結果。 所以。

不幸的是，並非您的所有回答都能奏效。 讓我們准備測試環境：

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;

現在，讓我們為每個訂單選擇“最后/最大狀態”，讓我們運行簡單的查詢：

 select id_order, max(date_add) as MaxDate
 from `order_history`
 group by id_order

這給了我們正確的結果，目前還沒有火箭科學：

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

現在為簡單起見，以免更改查詢以獲取Last State不等於3的 Orders。

我們期望獲得id_order = 3的一行結果 。

因此，讓我們測試一下我們的查詢：

通過RedFilter作出查詢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

結果：

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

結果：

id_order
--------
1
2

所以那不是真的

任何建議表示贊賞。

編輯

感謝Andriy M.評論，我們有適當的解決方案 。 它是對Tom H的修改。query的所有外觀如下：

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

編輯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

結果：

id_order
--------
3

所以這是真的