Select row with the earliest date - sql

Question

I've got a query in which I am returning multiple rows for a given event, for example:
ID | DateTime | String
1017436 | 2013-09-13 05:19:20.000 | Hello
1017436 | 2013-09-13 11:49:00.000 | World
I want the result to contain only the earliest occurrences of the row for any given ID, but am running into trouble.
I originally, thought a query like this would be the answer:
; WITH cte AS
( SELECT *,
rn = ROW_NUMBER() OVER (PARTITION BY ixBug ORDER BY dt)
FROM dbo.BugEvent
)
SELECT ixBug, dt, s
FROM cte
WHERE
ixBug IN (SELECT Bug.ixBug
FROM
dbo.Bug
JOIN
dbo.Mailbox ON Mailbox.ixMailbox = Bug.ixMailbox
WHERE
ixBug = '1017436'
AND
YEAR(dtOpened) >= '2013'
AND
MONTH(dtOpened) = '09'
AND
sOriginalTitle NOT LIKE '\[web\]%' ESCAPE '\'
AND
dbo.Bug.ixProject = (SELECT ixProject
FROM dbo.Project
WHERE sProject = 'Support')
AND
dbo.Bug.ixCategory = (SELECT ixCategory
FROM dbo.Category
WHERE sCategory = '.inquiry')
AND
Bug.ixBug NOT IN(SELECT Bug.ixBug
FROM
dbo.Bug
JOIN
dbo.Mailbox ON Mailbox.ixMailbox = Bug.ixMailbox
WHERE
YEAR(dtOpened) >= '2013'
AND
MONTH(dtOpened) <= '09'
AND
sOriginalTitle LIKE '\[web\]%' ESCAPE '\'
AND
dbo.Bug.ixProject = (SELECT ixProject
FROM dbo.Project
WHERE sProject = 'Support')
AND
dbo.Bug.ixCategory = (SELECT ixCategory
FROM dbo.Category
WHERE sCategory = '.inquiry')))
AND
sVerb = 'Incoming Email';
But, for some reason the result keeps both rows.

Answer 1

You can use window functions for this, either ROW_NUMBER() or MIN(). The idea is to partition the rows by the ID - OVER (PARTITION BY id) - and then either assign row numbers (ordered by the datetime) or find the minimum datetime per ID.
Solution with ROW_NUMBER():
; WITH cte AS
( SELECT id, datetime, string,
rn = ROW_NUMBER() OVER (PARTITION BY id ORDER BY datetime)
FROM tableX
)
SELECT id, datetime, string
FROM cte
WHERE rn = 1 ;
and with MIN():
; WITH cte AS
( SELECT id, datetime, string,
min_datetime = MIN(datetime) OVER (PARTITION BY id)
FROM tableX
)
SELECT id, datetime, string
FROM cte
WHERE datetime = min_datetime ;
The second version has slightly different behaviour. If there are two (or more) rows with exactly same datetime for an ID, they will be both in the result.

Answer 2

You can use ROW_NUMBER() which generates sequential number that you can filter with.
SELECT ID, DateTime, String
FROM
(
SELECT ID, DateTime, String,
ROW_NUMBER() OVER (PARTITION BY ID ORDER BY DateTime) RN
FROM tableName
) a
WHERE RN = 1
SQLFiddle Demo

Answer 3

Select the min date for each ID in a common table expression CTE, then join back to it.
with minDates (id, date)
as
(
select id, min(date) as date from YourTable
group by id
)
select yt.*
from YourTable yt
inner join minDates md on yt.id = md.id and yt.date = md.date

Answer 4

select top 1 id, date_time, string
from table
where id = ?
order by date_time

Answer 5

Select
ID
, DateTime
, String
From MyTable t1
Where t1.DateTime =
(
Select Min(t2.DateTime)
From MyTable t2
Where td.ID = t1.ID
)
Use a subquery to get the min DateTime for each ID group, then use that to qualify which of the IDs you pull. In case of tie, though, you will get both. If the IDs are in time sequence, you could avoid this by using the ID rather than the DateTime as the selector.