使用 GROUP BY 仅显示每个用户的最新更新行

Question

我一直无法弄清楚执行GROUP BY 的语法以仅显示每个用户（db_UserId）具有最新ups.db_LastUpdate 的行。

SELECT up.db_FirstName, up.db_LastName, up.db_UserId, ups.db_Initials, ups.db_LastUpdate 
FROM tblUserProfile up
    JOIN tblUserSel ups
        ON ups.db_Code = up.db_UserId
WHERE ups.db_UserTech = 'U'

输出（会有几百个用户，但你明白了）：

Jeff | Ledger | 1-34 | JL | 2015-08-11
Jeff | Ledger | 1-34 | DBC | 2015-06-06
Jeff | Ledger | 1-34 | YX | 2015-08-01
John | Barker | 1-26 | JR | 2015-04-04
John | Barker | 1-26 | YY | 2015-02-18
John | Barker | 1-26 | FF | 2015-11-14

也许像GROUP BY ups.dbUserId, MAX(db_LastUpdate)

感谢您的帮助

Answer 1

如果您的表有唯一的 ID 列，我通常会这样处理这种情况：

WITH LastEdit AS (
    SELECT ups.db_Code, ups.db_Initials, ups.db_LastUpdate
    FROM tblUserSel ups
    WHERE ups.db_UserTech = 'U' AND ups.ID = (
        SELECT TOP 1 ID
        FROM tblUserSel upsn
        WHERE ups.db_Initials = upsn.db_Initials
        ORDER BY upsn.db_LastUpdate DESC
    )
)
SELECT up.db_FirstName, up.db_LastName, up.db_UserId, le.db_Initials, le.db_LastUpdate 
FROM tblUserProfile up
    INNER JOIN LastEdit le
        ON le.db_Code = up.db_UserId

Answer 2

使用ROW_NUMBER:

;WITH CTE AS
(
    SELECT  up.db_FirstName, 
            up.db_LastName, 
            up.db_UserId, 
            ups.db_Initials, 
            ups.db_LastUpdate,
            RN = ROW_NUMBER() OVER(PARTITION BY up.db_UserId ORDER BY ups.db_LastUpdate DESC)
    FROM tblUserProfile up
    INNER JOIN tblUserSel ups
            ON ups.db_Code = up.db_UserId
    WHERE ups.db_UserTech = 'U'
)
SELECT *
FROM CTE
WHERE RN = 1;

正如 cmets 中所指出的，您可以使用 MAX 然后加入您的表格：

;WITH CTE AS
(
    SELECT  up.db_UserId, 
            MAX(ups.db_LastUpdate) MaxLastUpdate
    FROM tblUserProfile up
    INNER JOIN tblUserSel ups
            ON ups.db_Code = up.db_UserId
    WHERE ups.db_UserTech = 'U'
    GROUP BY up.db_UserId
)
SELECT B.*
FROM CTE A
INNER JOIN tblUserSel B
    ON A.db_UserId = B.db_Code
    AND A.MaxLastUpdate = B.db_LastUpdate;

但您需要知道，如果存在同一用户的同一日期的行，您将获得这 2 行作为结果。