Get max min per group in sequence

Question

I have a list like this below with desired result, please help.
The X are GPS cordinates, the time is the GPS times
X    time
A    5    
B    6
B    7
C    8
C    9
A    10
B    11
The result must be:
X    min    max
A    5    5
B    6    7
C    8    9
A    10    10
B    11    11

Answer 1

Here is a slightly diff solution, but you need SS 2012 or greater.
The idea is expressed in each CTE.
- Identify each new gps coordinate in time sequence per each car (new_gps).
- Count how many new_gps you have for each row in specific partition and windows frame (grp_helper).
- Aggregate to find min / max for each group (car, x, grp_helper).
- Bring next minTime
- APPLY to split the rows waiting and then moving
SET NOCOUNT ON;
USE tempdb;
GO
Create table #test (Car int, X Char(1), [Time] int)
INSERT INTO #test
VALUES
( 11, 'A', 5 ),
( 11, 'B', 6 ),
( 11, 'B', 7 ),
( 11, 'B', 8 ),
( 11, 'C', 9 ),
( 11, 'C', 10 ),
( 11, 'C', 11 ),
( 11, 'A', 12 ),
( 11, 'B', 13 );
-- ( 12, 'A', 15 ),
-- ( 12, 'B', 16 ),
-- ( 12, 'B', 17 ),
-- ( 12, 'B', 18 );
WITH C1 AS (
SELECT
Car,
X,
[Time],
CASE
WHEN X = LAG(X, 1, NULL) OVER(PARTITION BY Car ORDER BY [Time]) THEN 0 ELSE 1
END AS new_gps
FROM
#test
, C2 AS (
SELECT
Car,
X,
[Time],
SUM(new_gps) OVER(
PARTITION BY Car
ORDER BY [Time]
ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
) AS grp_helper
FROM
C1
Agg AS (
SELECT
Car,
X,
MIN([Time]) AS minTime,
MAX([Time]) AS maxTime,
COUNT(*) AS cnt
FROM
C2
GROUP BY
Car,
X,
grp_helper
, C3 AS (
SELECT
Car,
X,
minTime,
maxTime,
cnt,
LEAD(minTime, 1, maxTime) OVER(PARTITION BY Car ORDER BY minTime) AS nxt_minTime
FROM
Agg
SELECT
T.Car,
T.X,
T.minTime,
T.maxTime,
T.moving
FROM
C3
CROSS APPLY
SELECT C3.Car, C3.X, C3.minTime, C3.maxTime, 1 AS moving
WHERE cnt = 1
UNION ALL
SELECT C3.Car, C3.X, C3.minTime, C3.maxTime, 0 AS moving
WHERE cnt > 1
UNION ALL
SELECT C3.Car, C3.X, C3.maxTime, C3.nxt_minTime, 1 AS moving
WHERE cnt > 1
) AS T
ORDER BY
T.Car,
T.minTime;
GO
DROP TABLE #test;
GO
AMB
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