Question

使用以下数据，如何编写SQL Server查询以仅获取下面标有“X”的记录？基本上我希望记录具有相似的每组记录的MAX日期。下面有两组;第一个只有KEY1中的数据，第二个在KEY1和KEY2中都有数据。我尝试使用OVER语句，但可以得到我需要的东西。感谢。

DATE ------ ------ KEY1 KEY2

01-JAN ...... ABC .......... NULL

02-Jan ...... def ........... NULL'X'

12 ......二月ABC ......... 123

14-Feb ...... abc ......... 456'X'

所以这里是一个更真实的数据集问题的重新措辞。

Row     Date     Rate     Key1     Key2     Key3

1       01-Jan   150      12345  
2       05-Jan   155      12345     
3       01-Jan   160      12345    J100 
4       07-Feb   170      12345    J100 
5       09-Jan   170      12345    K200 
6       14-Jan   150      12345    J100     ABC
7       23-Jan   175      12345    J100     ABC

我想要获得的行是2,4,5和7，因为它们分别代表三个关键列的每个唯一组合的最大日期。希望更有意义。感谢。

Answer 1

select distinct LAST_VALUE(date) over (partition by key1, key2, key3 order by date) date,
                LAST_VALUE(rate) over (partition by key1, key2, key3 order by date) rate,
                LAST_VALUE(key1) over (partition by key1, key2, key3 order by date) key1,
                LAST_VALUE(key2) over (partition by key1, key2, key3 order by date) key2,
                LAST_VALUE(key3) over (partition by key1, key2, key3 order by date) key3
from t1

请注意，这只适用于SQL Server 2012。

在其他版本

  select t1.* 
    from t1, (select key1, key2, key3, max(date) date from t1 group by key1, key2, key3)) t2
where t2.key1 = t1.key1 and t2.key2 = t1.key2 and t2.key3 = t1.key3 and t2.date = t1.date

Answer 2

这很难看，但它对我有用。

select r.rate, t1.* from tRate r join (select max(r.date) date, r.key1, r.key2, r.key3 from tRate r join (select distinct key1+key2+key3 newKey from tRate) t2 on r.key1+r.key2+r.key3 = t2.newKey group by r.key1, r.key2, r.key3) t1 on r.date = t1.date and r.key1 = t1.key1 and r.key2 = t1.key2 and r.key3 = t1.key3

如何通过SQL Server查询获取以下数据

2 个答案: