Select 最大值和按日期分组的相应行 id

Question

I have a (mssql) ranking list table which has an id , a username , a date and a score attribute.

id | username | date | score
----------------------------
1      joe    01/2020  200
2      bob    01/2020  300
3      max    02/2020  350
4      jane   02/2020  300
5      bob    02/2020  250
6      joe    03/2020  150

What I'm trying to achieve is to select the highest score grouped by date with the corresponding id and username, which would produce a result like the following:

id | username | date | score
----------------------------
2      bob    01/2020  300
3      max    02/2020  350
6      joe    03/2020  150

I'm by no means a database guy and I failed miserably!

Can someone help me with this? Thanks!

Edit: I don't care about ties at this point.

Answer 1

if your database supports window functions:

Select * from (Select t.*,
row_number() OVER (Partition by date order by score desc)  rn
from table t) where rn = 1;


+----+----------+---------+-------+----+
| id | username | dte     | score | rn |
+----+----------+---------+-------+----+
|  2 | bob      | 01/2020 |   300 |  1 |
+----+----------+---------+-------+----+
|  3 | max      | 02/2020 |   350 |  1 |
+----+----------+---------+-------+----+
|  6 | joe      | 03/2020 |   150 |  1 |
+----+----------+---------+-------+----+

Answer 2

Try like below

select a.* form table_name a
where a.score=( select max(score) from table_name b where a.date=b.date )