get only last row in each day's multiple entries in TSQL
I have a table, something like:
Id Name 开发者_开发知识库 EnteredOn Percentage
`````````````````````````````````````````````````````````````
01 person1 2011-03-09 17:29:35.683 56.29
02 person1 2011-03-09 17:29:35.731 76.29
03 person1 2011-03-09 18:15:78.683 56.29
04 person1 2011-03-10 17:29:35.683 56.29
05 person1 2011-03-10 16:29:31.683 56.29
06 person1 2011-03-11 17:29:35.683 56.29
To summarize the above table, there are three rows for day 09, and two rows for day 10.
Now, I just want to select the latest row - one single row - per day.
(one row for 9, one for 10 and the one for 11)I cannot use distinct because of the timestamp. I cant group and use:
CAST(CONVERT(FLOAT, EnteredOn) AS INT)
because when I select EnteredOn field, it complaints that its not grouped. I cant combine distinct(cast..date...)
because I cant get the right syntax.
How can I select - only Name, EnteredOn, Percentage fields with distinct to each day?
many thanks in advance.
;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
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
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
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)
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