Question

这是我的样本数据集

id   thing  datetime      price
----|-----|------------|----------
1   | A   | 04/03/2009 |  399 
2   | B   | 04/03/2009 |  244
5   | C   | 04/03/2009 |  555
3   | D   | 04/03/2009 |  300
4   | A   | 03/03/2009 |  200
6   | B   | 03/03/2009 |  500
7   | C   | 24/12/2008 |  600
8   | D   | 01/01/2009 |  700
9   | A   | 01/03/2009 |  250
10  | B   | 01/03/2009 |  400
11  | C   | 12/12/2008 |  300
12  | D   | 20/01/2008 |  600

我需要从最大日期到每行的最后价格获取所有行。返回行应该像

id  thing  datetime      price  last price
---|-----|------------|--------|-----------
1  | A   | 04/03/2009 |  399   | 200
2  | B   | 04/03/2009 |  244   | 500
5  | C   | 04/03/2009 |  555   | 600
3  | D   | 04/03/2009 |  300   | 700

我可以在单个Sql中获得这个吗？在sql语句和sqlalchemy中需要这个 Sry任何错误，在此先感谢：）

Answer 1

我认为窗口功能可以解决您的问题

select id, thing, datetime, price 
from (
select id, thing, datetime, price,  rank()over(order by datetime desc ) rk
from yourTable )
where rk = 1;

Answer 2

这可能不是最好的方法，但它是SQL SERVER的解决方案，请查看

    with MY_TABLE as(

SELECT 1   id, 'A' thing , cast('20090403' as date) datetime,  399 price UNION ALL
SELECT 2   , 'B'   , '20090403' ,  244 UNION ALL
SELECT 5   , 'C'   , '20090403' ,  555 UNION ALL
SELECT 3   , 'D'   , '20090304' ,  300 UNION ALL
SELECT 4   , 'A'   , '20090303' ,  200 UNION ALL
SELECT 6   , 'B'   , '20090303' ,  500 UNION ALL
SELECT 7   , 'C'   , '20081224' ,  600 UNION ALL
SELECT 8   , 'D'   , '20090301' ,  700 UNION ALL
SELECT 9   , 'A'   , '20090301' ,  250 UNION ALL
SELECT 10  , 'B'   , '20090301' ,  400 UNION ALL
SELECT 11  , 'C'   , '20081212' ,  300 UNION ALL
SELECT 12  , 'D'   , '20080120' ,  600 

)



SELECT id,
        t1.thing,
        DATETIME,
        Y.price
FROM MY_TABLE t1
INNER JOIN (SELECT thing,price,rank()over(partition by thing order by datetime desc) as last_price
            from MY_TABLE) as Y on Y.thing= t1.thing and Y.last_price =2

WHERE datetime = (SELECT MAX(datetime) 
                  FROM MY_TABLE t2 
                  WHERE t1.thing = t2.thing)

如何使用MAX（日期）选择行并以前一个价格连接每一行

2 个答案: