get only last row in each day's multiple entries in TSQL

Question

I have a table, something like:

Id        Name        EnteredOn                    Percentage
`````````````````````````````````````````````````````````````
         


        

Answer 1

;with cte as
(
  select
    *,
    row_number() over(partition by datediff(d, 0, EnteredOn) order by EnteredOn desc) as rn 
  from YourTable
)
select *
from cte  
where rn = 1

Answer 2

I would suggest one more trick here:

select top 1 with ties 
    Name, EnteredOn, Percentage
from YourTable
order by row_number() over(partition by datediff(d, 0, EnteredOn) order by Name, EnteredOn desc)

Answer 3

SELECT Name, EnteredOn, Percentage
FROM (  SELECT *, ROW_NUMBER() OVER(PARTITION BY CONVERT(VARCHAR(8),EnteredOn,112) ORDER BY EnteredOn DESC) Corr
        FROM YourTable) A
WHERE Corr = 1

Answer 4

1 row/day:

SELECT t1.Name, t1.EnteredOn, t1.Percentage
  FROM table t1
  JOIN (SELECT MAX(EnteredOn) Max_EnteredOn_By_Day
          FROM table 
         GROUP BY convert(varchar, EnteredOn, 112)) t2
  ON t1.EnteredOn = t2.Max_EnteredOn_By_Day

1 row/person/day:

SELECT t1.Name, t1.EnteredOn, t1.Percentage
  FROM table t1
  JOIN (SELECT Name, MAX(EnteredOn) Max_EnteredOn_By_Day
          FROM table 
         GROUP BY Name, convert(varchar, EnteredOn, 112)) t2
  ON t1.Name = t2.Name
 AND t1.EnteredOn = t2.Max_EnteredOn_By_Day