counting most recent consecutive rows with like data using tabibitosan - sql

Question

My project is using an Oracle SQL database. I have a historical table that appends task status on a weekly basis, and am attempting to query the number of weeks a task that is currently off track has been off track. Here's an example excerpt from my source historical table:
ID WEEK ON_TRACK
1 1 N
1 2 Y
1 3 N
1 4 N
1 5 N
2 1 N
2 2 N
2 3 Y
2 4 Y
2 5 N
3 1 N
3 2 N
3 3 Y
3 4 Y
3 5 Y
I'm looking to return the count of consecutive "N" values in ON_TRACK starting backwards from the latest append. For the above example data, I'd like the query to return:
ID WKS_OFF_TRACK
1 3
2 1
3 0
I've done some research, and it looks like the Tabibitosan method is the most logical approach, and I've found ample examples to give the max consecutive values that match 1 criteria, but I'm having trouble tweaking to return the most recent consecutive values that match 2 criteria (ID and ON_TRACK).
Here's what I have so far
--this step creates a temp table with unique IDs for each weekly append to the historical table, and a 1 (if ON_TRACK = N) or 0 (if ON_TRACK = Y). This results in the expected info.
WITH HIST_TBL AS (
SELECT DISTINCT(ID),
CASE ON_TRACK
WHEN 'N' THEN 1
ELSE 0
END AS OFF_TRACK,
WEEK
FROM SOURCE_HISTORICAL_TBL
ORDER BY ID,WEEK DESC)
-- end of temp table
--this is where Im struggling I want one line per project number, and the sum of the latest string of 1s (weeks the task has been off track), until a 0 is reached.
SELECT ID,
SUM(OFF_TRACK) AS WKS_OFF_TRACK
FROM (SELECT WEEK,
ID,
OFF_TRACK,
ROW_NUMBER() OVER (ORDER BY WEEK DESC) - ROW_NUMBER() OVER
(PARTITION BY ID,OFF_TRACK ORDER BY WEEK DESC) GRP
FROM HIST_TBL)
GROUP BY ID, GRP
ORDER BY ID;
This code results in the a cumulative sum of all weeks each project has been off track, which for my example data would be:
ID WKS_OFF_TRACK
1 4
2 3
3 2
Any ideas where I'm going wrong?

Answer 1

Here is one method that assumes people were "on track" at some point in time:
select sht.id, count(*)
from SOURCE_HISTORICAL_TBL sht
where sht.week > (select max(sht2.week)
from SOURCE_HISTORICAL_TBL sht2
where sht2.id = sht.id and sht2.on_track = 'Y'
)
group by sht.id;
Otherwise, you need one more condition:
select sht.id, count(*)
from SOURCE_HISTORICAL_TBL sht
where sht.week > (select max(sht2.week)
from SOURCE_HISTORICAL_TBL sht2
where sht2.id = sht.id and sht2.on_track = 'Y'
) or
not exists (select 1
from SOURCE_HISTORICAL_TBL sht2
where sht2.id = sht.id and sht2.on_track = 'Y'
)
group by sht.id;
You can also phrase these as analytic functions:
select id,
sum(case when week > max_week_y or max_week_y is null then 1 else 0 end) as max_off_track
from (select sht.*,
max(case when on_track = 'Y' then week end) over (partition by id) as max_week_y
from SOURCE_HISTORICAL_TBL sht
) sht
group by id;
Note that this version will return 0s for people currently on track.

Answer 2

You can do it in a single table scan:
SQL Fiddle
Oracle 11g R2 Schema Setup:
CREATE TABLE SOURCE_HISTORICAL_TBL ( ID, WEEK, ON_TRACK ) AS
SELECT 1, 1, 'N' FROM DUAL UNION ALL
SELECT 1, 2, 'Y' FROM DUAL UNION ALL
SELECT 1, 3, 'N' FROM DUAL UNION ALL
SELECT 1, 4, 'N' FROM DUAL UNION ALL
SELECT 1, 5, 'N' FROM DUAL UNION ALL
SELECT 2, 1, 'N' FROM DUAL UNION ALL
SELECT 2, 2, 'N' FROM DUAL UNION ALL
SELECT 2, 3, 'Y' FROM DUAL UNION ALL
SELECT 2, 4, 'Y' FROM DUAL UNION ALL
SELECT 2, 5, 'N' FROM DUAL UNION ALL
SELECT 3, 1, 'N' FROM DUAL UNION ALL
SELECT 3, 2, 'N' FROM DUAL UNION ALL
SELECT 3, 3, 'Y' FROM DUAL UNION ALL
SELECT 3, 4, 'Y' FROM DUAL UNION ALL
SELECT 3, 5, 'Y' FROM DUAL UNION ALL
SELECT 4, 1, 'N' FROM DUAL UNION ALL
SELECT 5, 1, 'Y' FROM DUAL;
Query 1:
SELECT ID,
GREATEST(
COALESCE( MAX( CASE ON_TRACK WHEN 'N' THEN WEEK END ), 0 )
- COALESCE( MAX( CASE ON_TRACK WHEN 'Y' THEN WEEK END ), 0 ),
0
) AS weeks
FROM SOURCE_HISTORICAL_TBL
GROUP BY id
ORDER BY id
Results:
| ID | WEEKS |
|----|-------|
| 1 | 3 |
| 2 | 1 |
| 3 | 0 |
| 4 | 1 |
| 5 | 0 |