从聚合函数结果等栏目包括数据

Question

client_ref matter_suffix date_opened 
1   1    1983-11-15 00:00:00.000 
1   1    1983-11-15 00:00:00.000 
1   6    2002-11-18 00:00:00.000 
1   7    2005-08-01 00:00:00.000 
1   9    2008-07-01 00:00:00.000 
1   10    2008-08-22 00:00:00.000 
2   1    1983-11-15 00:00:00.000 
2   2    1992-04-21 00:00:00.000 
3   1    1983-11-15 00:00:00.000 
3   2    1987-02-26 00:00:00.000 
4   1    1989-01-07 00:00:00.000 
4   2    1987-03-15 00:00:00.000 

我上面的表，我只是想返回打开每个客户端的最近的事情，在下面的格式：

client_ref matter_suffix Most Recent 
1   10     2008-08-22 00:00:00.000 
2   2     1992-04-21 00:00:00.000 
3   2     1987-02-26 00:00:00.000 
4   1     1989-01-07 00:00:00.000 

我可以执行一个非常简单的查询来返回最近的（如下所示），但每当我尝试包含matter_suffix数据（必要）时，我都会遇到问题。

在此先感谢。

select client_ref,max (Date_opened)[Most Recent] from archive a 
group by client_ref 
order by client_ref 

Answer 1

在SQL 2012有方便的功能，使它更容易些，但在2008年SQL你需要做旧的方式：

找到最新：

SELECT client_ref,MAX(date_opened) last_opened 
FROM YourTable 
GROUP BY client_ref 

现在加入到背：

SELECT client_ref,matter_suffix, date_opened 
FROM YourTable YT 
INNER JOIN 
(
SELECT client_ref,MAX(date_opened) last_opened 
FROM YourTable 
GROUP BY client_ref 
) MR 
ON YT.client_ref = MR.client_ref 
AND YT.date_opened = MR.last_opened 

Answer 2

这不工作？

select client_ref,matter_suffix,max (Date_opened)[Most Recent] from archive a 
group by client_ref,matter_suffix 
order by client_ref 

Answer 3

可以使用ROW_NUMBER功能，让每client_ref的最新记录：

SELECT client_ref, matter_suffix, Date_opened 
FROM ( SELECT client_ref, 
        matter_suffix, 
        Date_opened, 
        RowNumber = ROW_NUMBER() OVER(PARTITION BY client_ref 
                ORDER BY Date_opened DESC) 
      FROM archive a 
     ) a 
WHERE RowNumber = 1; 

如果您想返回所有行有一个以上的行具有相同的最大敞开日期，那么你可以使用RANK：

SELECT client_ref, matter_suffix, Date_opened 
FROM ( SELECT client_ref, 
        matter_suffix, 
        Date_opened, 
        RowNumber = RANK() OVER(PARTITION BY client_ref 
                ORDER BY Date_opened DESC) 
      FROM archive a 
     ) a 
WHERE RowNumber = 1; 

Answer 4

select client_ref,max (matter_suffix),RN = ROW_NUMBER()OVER(PARTITION BY matter_suffix ORDER BY Date_opened desc) from archive a 
WHERE RN = 1 
group by client_ref,Date_opened 
order by client_ref,Date_opened