Oracle query - select date and time - Overlapping - sql

Question

oralce query Date and time with overlapping
ID startdate enddate hours
a124 10/10/2019 07:30:00 10/10/2019 11:30:00 4
a124 10/10/2019 07:00:00 10/10/2019 15:10:00 8.17
bc24 10/10/2019 07:30:00 10/10/2019 11:30:00 4
bc24 10/10/2019 10:30:00 10/10/2019 15:30:00 5
er67 10/10/2019 09:30:00 10/10/2019 11:30:00 2
er67 10/10/2019 15:30:00 10/10/2019 16:30:00 1
Desired Output :
ID startdate enddate hours
a124 10/10/2019 07:00:00 10/10/2019 15:10:00 8.17
bc24 10/10/2019 07:30:00 10/10/2019 15:30:00 8
er67 10/10/2019 09:30:00 10/10/2019 11:30:00 2
er67 10/10/2019 15:30:00 10/10/2019 16:30:00 1

Answer 1

I would approach this using lag(), a cumulative sum() and aggregation. Here is a step by step explanation.
First, you can use lag() to recover the previous start and end date for the same id:
select
t.*,
lag(startdate) over(partition by id order by startdate) lagstartdate,
lag(enddate) over(partition by id order by startdate) lagenddate
from mytable t
ID | STARTDATE | ENDDATE | HOURS | LAGSTARTDATE | LAGENDDATE
:--- | :------------------ | :------------------ | ----: | :------------------ | :------------------
a124 | 2019-10-10 07:00:00 | 2019-10-10 15:10:00 | 8.17 | null | null
a124 | 2019-10-10 07:30:00 | 2019-10-10 11:30:00 | 4 | 2019-10-10 07:00:00 | 2019-10-10 15:10:00
bc24 | 2019-10-10 07:30:00 | 2019-10-10 11:30:00 | 4 | null | null
bc24 | 2019-10-10 10:30:00 | 2019-10-10 15:30:00 | 5 | 2019-10-10 07:30:00 | 2019-10-10 11:30:00
er67 | 2019-10-10 09:30:00 | 2019-10-10 11:30:00 | 2 | null | null
er67 | 2019-10-10 15:30:00 | 2019-10-10 16:30:00 | 1 | 2019-10-10 09:30:00 | 2019-10-10 11:30:00
Then, you can set up the cumulative sum to slit records having the same id within groups (that will later on be aggregated). When the dates do not overlap, then a new group starts:
select
t.*,
sum(
case when startdate <= lagenddate or enddate <= lagstartdate
then 0
else 1
end
) over(partition by id order by startdate) grp
from (
select
t.*,
lag(startdate) over(partition by id order by startdate) lagstartdate,
lag(enddate) over(partition by id order by startdate) lagenddate
from mytable t
) t
ID | STARTDATE | ENDDATE | HOURS | LAGSTARTDATE | LAGENDDATE | GRP
:--- | :------------------ | :------------------ | ----: | :------------------ | :------------------ | --:
a124 | 2019-10-10 07:00:00 | 2019-10-10 15:10:00 | 8.17 | null | null | 1
a124 | 2019-10-10 07:30:00 | 2019-10-10 11:30:00 | 4 | 2019-10-10 07:00:00 | 2019-10-10 15:10:00 | 1
bc24 | 2019-10-10 07:30:00 | 2019-10-10 11:30:00 | 4 | null | null | 1
bc24 | 2019-10-10 10:30:00 | 2019-10-10 15:30:00 | 5 | 2019-10-10 07:30:00 | 2019-10-10 11:30:00 | 1
er67 | 2019-10-10 09:30:00 | 2019-10-10 11:30:00 | 2 | null | null | 1
er67 | 2019-10-10 15:30:00 | 2019-10-10 16:30:00 | 1 | 2019-10-10 09:30:00 | 2019-10-10 11:30:00 | 2
Finally, you can group the records by id and grp: min() and max() give you the date range, then you can compute the date difference.
Final query:
select
id,
min(startdate) startdate,
max(enddate) enddate,
round((max(enddate) - min(startdate)) * 24, 2) hours
from (
select
t.*,
sum(
case when startdate <= lagenddate or enddate <= lagstartdate
then 0
else 1
end
) over(partition by id order by startdate) grp
from (
select
t.*,
lag(startdate) over(partition by id order by startdate) lagstartdate,
lag(enddate) over(partition by id order by startdate) lagenddate
from mytable t
) t
) t
group by id, grp
order by id, grp
ID | STARTDATE | ENDDATE | HOURS
:--- | :------------------ | :------------------ | ----:
a124 | 2019-10-10 07:00:00 | 2019-10-10 15:10:00 | 8.17
bc24 | 2019-10-10 07:30:00 | 2019-10-10 15:30:00 | 8
er67 | 2019-10-10 09:30:00 | 2019-10-10 11:30:00 | 2
er67 | 2019-10-10 15:30:00 | 2019-10-10 16:30:00 | 1
Demo on DB Fiddle

Answer 2

You can use analytical functions (LAG and SUM) as following to make groups of dates without gaps:
SQL> SELECT
2 ID,
3 MIN(STARTDATE),
4 MAX(ENDDATE),
5 ROUND(SUM(CASE
6 WHEN PREV_ENDDATE BETWEEN STARTDATE AND ENDDATE THEN ENDDATE - PREV_ENDDATE
7 ELSE ENDDATE - STARTDATE
8 END) * 24, 2) AS HOURS
9 FROM
10 (
11 SELECT
12 T.*,
13 SUM(CASE
14 WHEN T.PREV_ENDDATE < T.STARTDATE THEN 1
15 END) OVER(
16 PARTITION BY ID
17 ORDER BY
18 STARTDATE, ENDDATE
19 ) SM
20 FROM
21 (
22 SELECT
23 ID,
24 STARTDATE,
25 ENDDATE,
26 LAG(ENDDATE) OVER(
27 PARTITION BY ID
28 ORDER BY
29 STARTDATE, ENDDATE
30 ) AS PREV_ENDDATE
31 FROM
32 T TOUT
33 WHERE
34 NOT EXISTS (
35 SELECT
36 1
37 FROM
38 T TIN
39 WHERE
40 TIN.ID = TOUT.ID
41 AND TOUT.STARTDATE BETWEEN TIN.STARTDATE AND TIN.ENDDATE
42 AND TOUT.ENDDATE BETWEEN TIN.STARTDATE AND TIN.ENDDATE
43 AND TOUT.ROWID <> TIN.ROWID
44 )
45 ) T
46 )
47 GROUP BY
48 ID,
49 SM;
ID MIN(START MAX(ENDDA HOURS
---- --------- --------- ----------
a124 10-OCT-19 10-OCT-19 8.17
bc24 10-OCT-19 10-OCT-19 8
er67 10-OCT-19 10-OCT-19 1
er67 10-OCT-19 10-OCT-19 2
SQL>
Cheers!!