SQL MIN Datetime based on first occuranceof a value in another column - sql

Question

This is what I have
ID Name DateTime Value Group
1 Mark 1/1/2010 0 1
2 Mark 1/2/2010 1 1
3 Mark 1/3/2010 0 1
4 Mark 1/4/2010 0 2
40 Mark 1/5/2010 1 2
5 Mark 1/9/2010 1 2
6 Mark 1/6/2010 1 2
7 Kelly 1/1/2010 0 3
8 Kelly 1/2/2010 1 3
9 Kelly 1/3/2010 1 3
10 Nancy 1/4/2010 0 4
11 Nancy 1/5/2010 0 4
12 Nancy 1/6/2010 1 5
13 Nancy 1/7/2010 0 5
What I want is to get the rows per "name" per "group" with minimum datetime after the value becomes 1. From the above example, I would need to get
3 Mark 1/3/2010 0 1
6 Mark 1/6/2010 1 2
9 Kelly 1/3/2010 1 3
13 Nancy 1/7/2010 0 5

Answer 1

Based on the description of your rules, I believe the output will actually be a bit different since 2010-01-05 was the first DateTime where the Value = 1 for Group 2 for Mark.
ID Name DateTime Value Group
3 Mark 2010-01-03 0 1
6 Mark 2010-01-06 1 2
9 Kelly 2010-01-03 1 3
13 Nancy 2010-01-07 0 5
The below code will work as demonstrated in this SQLFiddle.
SELECT sub.ID
, sub.Name
, sub.[DateTime]
, sub.Value
, sub.[Group]
FROM
(SELECT t.ID
, t.Name
, t.[DateTime]
, t.Value
, t.[Group]
, SequentialOrder = ROW_NUMBER() OVER
(PARTITION BY t.Name, t.[Group]
ORDER BY t.[DateTime])
FROM Test t
JOIN
(SELECT Name
, [Group]
, MinimumDateTime = MIN([DateTime])
FROM Test
WHERE Value = 1
GROUP BY Name
, [Group]) mint
ON t.Name = mint.Name
AND t.[Group] = mint.[Group]
WHERE t.[DateTime] > mint.MinimumDateTime) sub
WHERE sub.SequentialOrder = 1
ORDER BY ID;

Answer 2

Below is my query and it goes on assumption that records are received in order of their dates
WITH TBL_1 AS
(
SELECT A.*, ROW_NUMBER() OVER(PARTITION BY NAME, GROUP ORDER BY DATE) AS RN
FROM TABLE
WHERE (NAME, GROUP) IN
(SELECT NAME, GROUP FROM TABLE WHERE VALUE = 1)
),
TBL_2 AS
(
SELECT * FROM TBL_1 WHERE VALUE = 1
),
TBL_3 AS
(
SELECT A.*
FROM TBL_1 AS A
INNER JOIN TBL_2 AS B
ON B.NAME = A.NAME
AND B.GROUP = A.GROUP
AND A.RN > B.RN
)
SELECT *
FROM TBL_3
WHERE (NAME, GROUP, DATE) IN
(SELECT NAME, GROUP, MIN(DATE) FROM TBL_3 GROUP BY NAME, GROUP)

Answer 3

In SQL Server 2012 you can do this:
SELECT * FROM (
SELECT DISTINCT
ID,
Name,
DateTime,
Value,
Gr,
LAG(ID) OVER (PARTITION BY Name, Gr ORDER BY DateTime) F
FROM (
SELECT
ID,
Name,
DateTime,
Value,
Gr,
CASE WHEN LAG(Value) OVER (PARTITION BY Name, Gr ORDER BY DateTime) = 1 THEN 1 ELSE 0 END F
FROM
T
) TT
WHERE F = 1
) TT WHERE F IS NULL
ORDER BY Gr, Name, DateTime
Fiddle: http://www.sqlfiddle.com/#!6/5a0fa2/19

Answer 4

using window functions:
with cte as (
select
*,
row_number() over(partition by [Group], Name order by [DateTime]) as rn,
dense_rank() over(order by [Group], Name) as rnk
from Table1
)
select c1.*
from cte as c1
inner join cte as c2 on c2.rn = c1.rn - 1 and c2.rnk = c1.rnk and c2.Value = 1
where
not exists (select * from cte as c3 where c3.rn <= c1.rn - 2 and c3.rnk = c1.rnk and c3.Value = 1)
or apply:
select t1.*
from Table1 as t1
cross apply (
select top 1 t2.Value, t2.DateTime
from Table1 as t2
where
t2.[Group] = t1.[Group] and t2.Name = t1.Name and
t2.[DateTime] < t1.[DateTime]
order by t2.[Datetime] desc
) as t2
where
t2.Value = 1 and
not exists (
select *
from Table1 as t3
where
t3.[Group] = t1.[Group] and t3.Name = t1.Name and
t3.[DateTime] < t2.[DateTime] and t3.Value = 1
)
sql fiddle demo
update forgot to mention that your output seems to be incorrect - there should id = 6 instead of 5 in second row (see sql fiddle).