oracle SQL counter restarts when time difference is > x - sql

Question

I want to create a new column in my query whereby it takes into account the difference of the current rows datetime - previous datetime. This column could be a counter where if the difference is <-100, it stays as 1, but once there difference is > -100, the column is 0.
Ideally then I would want to only pull in the rows that come after the last 0 record.
My query:
with products as (
select * from (
select distinct
ID,
UnixDateTime,
OrderNumber,
to_date('1970-01-01','YYYY-MM-DD') + numtodsinterval(UnixDateTime,'SECOND')+ 1/24 as "Date_Time"
from DB
where
(date '1970-01-01' + UnixDateTime * interval '1' second) + interval '1' hour
> sysdate - interval '2' day
)
),
prod_prev AS (
SELECT p.*,
lag("Date_Time")over(order by "Date_Time" ASC) as Previous_Time,
lag(UnixDateTime)over(order by "Date_Time" ASC) as UnixDateTime_Previous_Time "Date_Time") - "Date_Time" AS diff
FROM products p
),
run_sum AS (
SELECT p.*, "Date_Time"-Previous_Time as "Diff", UnixDateTime_Previous_Time-UnixDateTime AS "UnixDateTime_Diff"
FROM prod_prev p
)
SELECT * FROM run_sum
ORDER By UnixDateTime, "Date_Time" DESC
my query result from above query:
ID
UnixDateTime
OrderNumber
Date_Time
Previous_Time
diff
UnixDateTime_Diff
1
1662615688
100
08-SEP-2022 06:41:28
(null)
(null)
(null)
2
1662615752
100
08-SEP-2022 06:42:32
08-SEP-2022 06:41:28
0.00074
-64
3
1662615765
100
08-SEP-2022 06:42:45
008-SEP-2022 06:42:32
0.000150
-13
4
1662615859
100
08-SEP-2022 06:44:19
08-SEP-2022 06:42:45
0.001088
-128
5
1662615987
100
08-SEP-2022 06:46:27
08-SEP-2022 06:44:19
0.00148
-44
6
1662616031
100
08-SEP-2022 06:47:11
08-SEP-2022 06:46:27
0.00051
-36
the counter is the below example should be 1 if the UnixDateTime_Diff is < -100 and 0 if its >-100
then if I could only pull in records AFTER the most recent 0 record.

Answer 1

You use:
lag("Date_Time")over(order by "Date_Time" DESC)
And get the previous value when the values are ordered in DESCending order; this will get the previous higher value. If you want the previous lower value then either use:
lag("Date_Time") over (order by "Date_Time" ASC)
or
lead("Date_Time") over (order by "Date_Time" DESC)
If you want to perform row-by-row processing then, from Oracle 12, you can use MATCH_RECOGNIZE:
SELECT id,
unixdatetime,
ordernumber,
date_time,
next_unixdatetime,
next_unixdatetime - unixdatetime AS diff,
CASE cls
WHEN 'WITHIN_100' THEN 1
ELSE 0
END AS within_100
from (
select distinct
ID,
UnixDateTime,
OrderNumber,
TIMESTAMP '1970-01-01 00:00:00 UTC' + UnixDateTime * INTERVAL '1' SECOND
AS Date_Time
from DB
where TIMESTAMP '1970-01-01 00:00:00 UTC' + UnixDateTime * INTERVAL '1' SECOND
> SYSTIMESTAMP - INTERVAL '2' DAY
)
MATCH_RECOGNIZE(
ORDER BY unixdatetime
MEASURES
NEXT(unixdatetime) AS next_unixdatetime,
classifier() AS cls
ALL ROWS PER MATCH
PATTERN (within_100* any_row)
DEFINE
within_100 AS NEXT(unixdatetime) < unixdatetime + 100
) m
Which, for the sample data:
CREATE TABLE db (ID, UnixDateTime, OrderNumber) AS
SELECT 1, 1662615688, 100 FROM DUAL UNION ALL
SELECT 2, 1662615752, 100 FROM DUAL UNION ALL
SELECT 3, 1662615765, 100 FROM DUAL UNION ALL
SELECT 4, 1662615859, 100 FROM DUAL UNION ALL
SELECT 5, 1662615987, 100 FROM DUAL UNION ALL
SELECT 6, 1662616031, 100 FROM DUAL;
Outputs:
ID
UNIXDATETIME
ORDERNUMBER
DATE_TIME
NEXT_UNIXDATETIME
DIFF
WITHIN_100
1
1662615688
100
2022-09-08 05:41:28.000000000 UTC
1662615752
64
1
2
1662615752
100
2022-09-08 05:42:32.000000000 UTC
1662615765
13
1
3
1662615765
100
2022-09-08 05:42:45.000000000 UTC
1662615859
94
1
4
1662615859
100
2022-09-08 05:44:19.000000000 UTC
1662615987
128
0
5
1662615987
100
2022-09-08 05:46:27.000000000 UTC
1662616031
44
1
6
1662616031
100
2022-09-08 05:47:11.000000000 UTC
null
null
0
fiddle