寻找另一个最大的查询

Question

我有一个表的数据，看起来像的最大如下：

+--------+----------+------------+ 
| Client | Item No. | Serial No. | 
+--------+----------+------------+ 
| A  |  1 |   1 | 
| A  |  1 |   2 | 
| B  |  1 |   2 | 
| B  |  2 |   1 | 
| C  |  2 |   3 | 
| C  |  2 |   2 | 
| C  |  3 |   1 | 
| C  |  3 |   2 | 
| D  |  2 |   3 | 
| D  |  2 |   1 | 
| D  |  3 |   2 | 
| D  |  3 |   3 | 
| D  |  2 |   2 | 
| D  |  3 |   1 | 
+--------+----------+------------+ 

我正在寻找做的是找到为每个客户最高产品编号其次是最高序列号（以此顺序）。因此，对于上面的输出将是：

+--------+----------+------------+ 
| Client | Item No. | Serial No. | 
+--------+----------+------------+ 
| A  |  1 |   2 | 
| B  |  2 |   1 | 
| C  |  3 |   2 | 
| D  |  3 |   3 | 
+--------+----------+------------+ 

我想这需要一个嵌套MAX（）语句，首先得到MAX（项目编号）为每一个客户，并为那些具有多那么得到MAX（序列号）什么是写这个查询的有效方法？

Answer 1

也许这可以帮助：

SELECT t.client, 
     t.item, 
     MAX(t.serial) AS serial 
    FROM(SELECT client, 
       MAX(item) AS item 
     FROM your_table 
     GROUP 
      BY client 
    ) a 
    JOIN your_table t 
    ON a.client = t.client AND a.item = t.item 
GROUP 
    BY t.client, t.item 

如果你的DBMS支持窗口功能，另一种方法是：

SELECT client, item, serial 
    FROM(SELECT client, item, serial, 
       ROW_NUMBER() OVER (PARTITION BY client ORDER BY item DESC, serial DESC) rn 
     FROM your_table 
    ) 
WHERE rn = 1 

Answer 2

Postgres有，因为DISTINCT的真棒快捷

SELECT DISTINCT ON (Client) Client, Item, SerialNumber 
FROM some_table 
ORDER BY Item DESC, SerialNumber DESC -- DESC gets highest 

这只会为每个客户端抓取一条记录，而ORDER BY会确保它是您想要的。

我不知道如果任何其他DBMSs会这么容易做到这一点。

Answer 3

这里是解决方案：

WITH CTE AS (
SELECT Client, MAX(Item_No) AS Item_No 
FROM YourTable 
GROUP BY Client) 

SELECT a.Client,b.Item_No,MAX(Serial_No) AS Serial_No 
FROM CTE a 
JOIN YourTable b ON a.Client = b.Client AND a.Item_No = b.Item_No 
GROUP BY a.Client,b.Item_No