SQL Select Distinct Top5 with Order By and Where

I have a table where Employees can check assets and each of their checks is recorded in a table.

Now I want to extract the 5 last checked Asset_Ids for a given employee. He might have checked the same asset more than once, each check is recorded in a table -EmployeeLog-

Basically my table has 3 columns :

Employee_ID	DateChecked	Asset_ID
1	06/10/2021 10:56:22	1
1	06/10/2021 14:58:25	1
1	02/10/2021 13:56:22	2
1	06/10/2021 08:56:22	1
1	04/10/2021 03:56:22	1
1	06/10/2021 02:56:22	3
1	06/10/2021 15:56:22	2
1	03/10/2021 03:56:22	5
1	06/10/2021 03:56:22	5

I have below query which lists Asset_IDs by check date, but I have to apply distinct and top 5 to this list which I couldn’t, any help appreciated.

SELECT Asset_ID
FROM EmployeeLog
WHERE EmployeeID = @Emp_ID AND Asset_ID != '00000000-0000-0000-0000-000000000000'
ORDER BY EmployeeLog.DateChecked Desc

Answer

You can do a GROUP BY, and order by first (or last) date.

SELECT Asset_ID
FROM EmployeeLog
WHERE EmployeeID = @Emp_ID AND Asset_ID != '00000000-0000-0000-0000-000000000000'
GROUP BY Asset_ID
ORDER BY MIN(EmployeeLog.DateChecked) Desc

To get the first 5 rows only, do SELECT TOP 5 Asset_ID etc.