我有如下表:
-----------------------------------------------------------
ID oDate oName oItem oQty oRemarks
-----------------------------------------------------------
1 2016-01-01 A 001 2
2 2016-01-01 A 002 1 test
3 2016-01-01 B 001 3
4 2016-01-02 B 001 2
5 2016-01-02 C 001 2
6 2016-01-03 B 002 1
7 2016-01-03 B 001 4
ff.
我想每一个名字的最新记录。所以结果应该是这样的:
-----------------------------------------------------------
oDate oName oItem oQty oRemarks
-----------------------------------------------------------
2016-01-01 A 001 2
2016-01-01 A 002 1 test
2016-01-02 C 001 2
2016-01-03 B 002 1
2016-01-03 B 001 4
ff.
有谁知道怎么做得到这样的结果? 谢谢。
该rank
窗口子句允许,那么,根据一些分区排名行,然后你可以只选择顶部的:
SELECT oDate, oName, oItem, oQty, oRemarks
FROM (SELECT oDate, oName, oItem, oQty, oRemarks,
RANK() OVER (PARTITION BY oName ORDER BY oDate DESC) AS rk
FROM my_table) t
WHERE rk = 1
这是一个没有使用分析功能的通用查询。
SELECT a.*
FROM table1 a
INNER JOIN
(SELECT max(odate) modate,
oname,
oItem
FROM table1
GROUP BY oName,
oItem
)
b ON a.oname=b.oname
AND a.oitem=b.oitem
AND a.odate=b.modate
一个primary key
假设id
字段添加到表中,并使其自动递增,。然后,为了通过id
你会得到它。这是传统的方式。通过使用你的表,你只能通过oDate
订购。但应具有相同的日期多次,所以它也不会解决你的问题。
我认为你需要这样的查询:
SELECT *
FROM (SELECT *,
ROW_NUMBER() OVER (PARTITION BY oName ORDER BY oDate DESC) seq
FROM yourTable) t
WHERE (seq <= 2)
ORDER BY oDate;
你必须在以下使用ROW_NUMBER
:
select oDate, oName, oItem, oQty, oRemarks
from (
select *, row_number() over(partition by oName, oItem order by oDate desc) rn
from #t
)x
where rn = 1
order by oDate
OUTPUT
oDate oName oItem oQty oRemarks
2016-01-01 A 001 2
2016-01-01 A 002 1 test
2016-01-02 C 001 2
2016-01-03 B 001 4
2016-01-03 B 002 1