Get datetime delta gaps

Question

Finding datetime gaps:
we have a table with cars working from to dates:
rec    car    DateFrom        DateTo
1    11    2008-10-28 00:00:00    2008-11-05 23:59:59
2    11    2008-11-06 00:00:00    2008-11-08 23:00:00
3    11    2008-11-09 00:00:00    2008-11-05 23:59:59
4    22    2008-10-28 00:00:00    2008-10-29 20:00:00
5    22    2008-10-31 00:00:00    2008-11-05 23:59:59
result I'm lookin for is finding dates and delta time a car is not working:
car    gapFrom         gapTo            gapDelta[h]
11    2008-11-08 23:00:00    2008-11-09 00:00:00    1     (in hours)
22    2008-10-29 20:00:00    2008-10-31 00:00:00    28    

Answer 1

Hi b_blase,
Please see the below code to retrieve datetime delta gaps.
CREATE TABLE yourTable
rec INT,
car INT,
dateFrom DATETIME,
dateTo DATETIME
INSERT INTO yourTable VALUES(1,11,'2008-10-28 00:00:00','2008-11-06 00:00:00'),
(2,11,'2008-11-06 00:00:00','2008-11-08 23:00:00'),
(3,11,'2008-11-09 00:00:00','2008-11-10 00:00:00'),
(4,11,'2008-11-10 11:37:00','2008-11-11 00:00:00'),
(5,22,'2008-10-28 00:00:00','2008-10-29 20:00:00'),
(6,22,'2008-10-31 00:00:00','2008-11-05 00:00:00'),
(7,22,'2008-11-07 00:00:00','2008-11-11 00:00:00');
;WITH CTE AS
SELECT *,ROW_NUMBER() OVER(PARTITION BY car ORDER BY dateFrom) AS rn FROM yourTable
CTE2 AS
SELECT c1.car car,c2.dateTo gapFrom, c1.dateFrom gapTo
FROM CTE c1 LEFT JOIN CTE c2
ON c1.rn=c2.rn+1 AND c1.car=c2.car
WHERE c1.dateFrom <> c2.dateTo
SELECT *,CAST(ROUND(DATEDIFF(MINUTE,gapFrom,gapTo)/60.0,2) AS NUMERIC(20,1)) gapDelta FROM CTE2 ORDER BY car,gapTo;
DROP TABLE yourTable;
I just made a little modification on the dateTo column so that the
dateTo equaling to the dateFrom of next record can be identified as no gap. If you still prefer the dateTo valuing like ‘2008-11-06 23:59:59‘ as it was, you just modify the WHERE clause as below.
WHERE DATEDIFF(MINUTE,c2.dateTo,c1.dateFrom) <> 1
If you have any question, feel free to let me know.
Best regards,
Eric Zhang