通过SQL获取每个组的最新记录

我有下表：

我想得到每个名字的最新记录。所以结果应该是这样的：

SELECT *
FROM (SELECT *,
        ROW_NUMBER() OVER (PARTITION BY oName ORDER BY oDate DESC) seq
    FROM yourTable) t
WHERE (seq <= 2)
ORDER BY oDate; 

有人知道如何得到这个结果吗？

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谢谢。

我想您需要这样的查询：

SELECT *
FROM (SELECT *,
        ROW_NUMBER() OVER (PARTITION BY oName ORDER BY oDate DESC) seq
    FROM yourTable) t
WHERE (seq <= 2)
ORDER BY oDate; 

您必须在以下情况下使用行号：

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

输出

window子句允许您根据某些分区对行进行排序，然后只需选择最上面的行：

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

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向表中添加主键假定id字段并使其自动递增，。然后按id订购，您将获得它。这是传统的方式。通过使用表格，您只能按oDate订购。但是，同一个日期是多次的，所以这也不能解决你的问题

这是一个不使用分析函数的通用查询

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结果不是每个名称的最新记录，因为存在重复的名称。@Haminteu添加主键buddy它将解决您的问题，您可以按此进行订购。@androidGenX，您好。只需编辑表格。我的表上有标识栏。干杯。@Haminteu检查答案。可能是重复的

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   001     4                   
2016-01-03  B   002     1                   

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

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