Group By 以获得最小交易

Question

目前我有如下交易清单：

Front Office ID    Transaction ID    TradeDate    SettlementDate
10000              1234              2015-03-03   2015-03-04
10000              1235              2015-03-03   2015-06-17
10001              1232              2015-03-13   2015-03-18
10001              1231              2015-03-13   2015-06-17

我需要做的是找到交易日期和结算之间天数最短的每个前台 ID 的交易 ID。

我知道如何获取日期差异，但我对如何通过以下方式在组中工作感到困惑：

MIN(DATEDIFF(DD,FX.TradeDate,FX.SettleDate))

任何正确方向的帮助或指示都非常受欢迎。

谢谢。

Answer 1

需要先按Front Office ID分组，再按数据差异排序，即可得到min.

像这样：

SELECT id  
FROM tbl 
GROUP BY id
ORDER BY difference

Answer 2

如果你想 "find transaction ID's per Front Office ID which has the shortest number of days between trade date and settlement" 为什么还需要 GROUP BY？

SELECT TOP 1 t.[Transaction ID] 
FROM dbo.Transactions t
WHERE t.[Front Office ID] = @FrontOfficeID
ORDER BY DATEDIFF(DD,  t.TradeDate, t.SettleDate) ASC

如果您想为所有交易 ID 查找它，您可以使用 CTE + ROW_NUMBER:

WITH CTE AS
(
    SELECT t.[Transaction ID],  t.[Front Office ID], t.TradeDate, t.SettleDate
           RN = ROW_NUMBER() OVER (PARTITION BY [Front Office ID] 
                                   ORDER BY DATEDIFF(DD,  t.TradeDate, t.SettleDate) ASC)
    FROM dbo.Transactions t
)
SELECT [Transaction ID],  [Front Office ID], TradeDate, SettleDate
FROM CTE
WHERE RN = 1

如果您想要 return 所有在 TradeDate 和 SettleDate 之间具有最短时间跨度的交易 ID（在平局的情况下），请使用 DENSE_RANK 而不是 ROW_NUMBER。

Answer 3

您可以使用 ROW_NUMBER() 并在 [Front Office ID] 上执行 PARTITION BY。

;WITH CTE AS 
( 
    SELECT
        [Front Office ID] ,
        [Transaction ID],
        ROW_NUMBER()OVER( PARTITION BY [Front Office ID] ORDER BY DATEDIFF(DD,FX.TradeDate,FX.SettleDate) ASC) rn
    FROM FX
)
SELECT [Front Office ID] ,[Transaction ID]
FROM CTE WHERE rn = 1

Group By 以获得最小交易

Group By to get min transaction

tsql

group-by

sql-server-2008