Fetch the rows which have the Max value for a column for each distinct value of another column

select UserId,max(Date) over (partition by UserId) value from users; 

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select distinct sno,item_name,max(start_date) over(partition by sno),max(end_date) over(partition by sno),max(creation_date) over(partition by sno), max(last_modified_date) over(partition by sno) from uniq_select_records order by sno,item_name asc;

will given accurate results related to that link

Use the code:

select T.UserId,T.dt from (select UserId,max(dt) over (partition by UserId) as dt from t_users)T where T.dt=dt; 

This will retrieve the results, irrespective of duplicate values for UserId. If your UserId is unique, well it becomes more simple:

select UserId,max(dt) from t_users group by UserId; 

SELECT a.userid,a.values1,b.mm FROM table_name a,(SELECT userid,Max(date1)AS mm FROM table_name GROUP BY userid) b WHERE a.userid=b.userid AND a.DATE1=b.mm; 

Below query can work :

SELECT user_id, value, date , row_number() OVER (PARTITION BY user_id ORDER BY date desc) AS rn FROM table_name WHERE rn= 1 

SELECT a.* FROM user a INNER JOIN (SELECT userid,Max(date) AS date12 FROM user1 GROUP BY userid) b ON a.date=b.date12 AND a.userid=b.userid ORDER BY a.userid;