從聚合函數結果等欄目包括數據

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，我只是想返回打開每個客戶端的最近的事情，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 

我可以執行一個非常簡單的查詢來返回最近的（如下所示），但每當我嘗試包含matter_suffix數據（必要）時，我都會遇到問題。

在此先感謝。

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你需要做舊的方式：

找到最新：

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

現在加入到背：

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

這不工作？

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，讓每client_ref的最新記錄：

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個以上的行具有相同的最大敞開日期，那麼你可以使用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