sql - SQL Server - Selecting multiple maximum values matching two separate corresponding cells

Question

I am a beginner in SQL and been trying to obtain all of the maximum values as per the access strength and folder names, with allowing groups to the folders being displayed accordingly. I tried using MAX function but it was returning only 1 maximum result. Thank you in advance for any help or guidance of how t achieve the specified results.

Please see the jpg link attached for the top table and the desired outcome in the bottom table.

John

Answer 1

您可以使用RANK功能将您的记录按顺序排列，然后只选择前 1 个：

WITH CTE AS
(   SELECT  Folder_name,
            [User],
            Access,
            Group_Allowing,
            [Rank] = RANK() OVER(PARTITION BY Folder_Name, [User] ORDER BY Access DESC)
    FROM    T
)
SELECT  Folder_name,
        [User],
        Access,
        Group_Allowing
FROM    CTE
WHERE   [Rank] = 1;

该PARTITION BY子句类似于 GROUP BY，每个组的排名将从 0 重新开始。然后，顺序简单地指示了按什么顺序进行排名。

Answer 2

您可以使用“ORDER BY Access DESC”和“LIMIT [您想要的条目数]”，或者如果您只想获得最高数字，您可以使用“GROUP BY Access”，然后使用 max() 显示您的所有条目想

Answer 3

您需要将您的查询MAX()放入子查询JOIN中：

SELECT a.*
FROM YourTable a
JOIN (SELECT Folder_Name, [User], MAX(Access) Max_Access
      FROM YourTable
      GROUP BY Folder_Name, [User]
     )b
 ON a.Folder_Name = b.Folder_Name
  AND a.[User] = b.[User]
  AND a.Access = b.Max_Access