sqlite query comparing data in different rows - sqlite

Question

i have the following table where i have the date( not a primary key) and rating ('A' being the highest grade):
date rating
03-10-2010 C
03-09-2010 C
03-08-2010 B
03-07-2010 B
03-06-2010 B
03-05-2010 B
03-04-2010 A
I need to make a query where i compare the rating in order to return the result for each 'date'.
For example. considering the date 03-10-2010, i want to know when the last rating downgrade happened. if the downgrade was 1 day ago return '1' as result, if it was 2 days ago return '2' and if was older than 3 days return 0.
And i would do the same query for each date, getting an array with the results.
i'm stuck trying to do this and i have no more ideas how to do it. Anyone can help me please?
thanks.

Answer 1

You want the difference, in days, between the date of each record and the date of the record before the last downgrade.
When you have a specific record, the record before the last downgrade is the record that
has a higher rating than this record, and
has a lower date than this record, and
is the latest record of those.
In SQL, this can be done with a correlated subquery:
SELECT date,
rating,
(SELECT date
FROM MyTable AS downgrade
WHERE downgrade.date < MyTable.date
AND downgrade.rating < MyTable.rating
ORDER BY date DESC
LIMIT 1) AS downgrade_date
FROM MyTable
date rating downgrade_date
---------- ---------- ----------
2010-03-04 A
2010-03-05 B 2010-03-04
2010-03-06 B 2010-03-04
2010-03-07 B 2010-03-04
2010-03-08 B 2010-03-04
2010-03-09 C 2010-03-08
2010-03-10 C 2010-03-08
To compute the difference, convert the date into a numeric value.
You can then use this value for further computations:
SELECT date,
rating,
CASE
WHEN days <= 3 THEN days
ELSE 0
END AS whatever
FROM (SELECT date,
rating,
julianday(date) -
julianday((SELECT date
FROM MyTable AS downgrade
WHERE downgrade.date < MyTable.date
AND downgrade.rating < MyTable.rating
ORDER BY date DESC
LIMIT 1)) AS days
FROM MyTable)