Create a SQLite view where a row depends on the previous row

Question

I\'d like to create a view in SQLite where a field in one row depends on the value of a field in the previous row. I could do this in Oracle using the LAG analy

Answer 1

This should do the trick for every item (tested on SQLite):

SELECT 
    day
    ,price
    ,price - (SELECT t2.price 
              FROM mytable t2 
              WHERE 
                  t2.item = t1.item AND 
                  t2.day < t1.day      
              ORDER BY t2.day DESC
              LIMIT 1
             ) AS change
FROM mytable t1

This assumes the combination between day and item is unique. And the way it works is by taking all the values less than the given day, sorting descending and then LIMIT just the first value, simulating a LAG function.

For a LEAD behavior, just flip < to > and DESC to ASC.

Answer 2

Oracle equivalent is correct. Starting from SQLite 3.25.0 you could use LAG natively:

WITH mytable(ITEM,DAY,PRICE) AS (
    VALUES
    ('apple',  CAST('20110107' AS DATE),    1.25),
    ('orange', CAST('20110102' AS DATE),    1.00),
    ('apple',  CAST('20110101' AS DATE),    1.00),
    ('orange', CAST('20110103' AS DATE),    2.00),
    ('apple',  CAST('20110108' AS DATE),    2.00),
    ('apple',  CAST('20110110' AS DATE),    1.50)
)
SELECT day, price, price-LAG(price) OVER (ORDER BY day) AS change
FROM mytable
WHERE item = 'apple'
ORDER BY DAY;

Answer 3

Assuming that you don't delete this will work:

SELECT t2.DAY, t2.price, t2.price-t1.price 
FROM TABLENAME t1, TABLENAME t2 
WHERE t1.rowid=t2.rowid-1

This works because every row has its own rowid even if you dont specify it in the CREATE statement.

If you do delete, it becomes:

SELECT t2.day, t2.price, t2.price-t1.price 
FROM 
     (SELECT l1.day, l1.price, 
          (SELECT COUNT(*) 
          FROM TABLENAME l2 
          WHERE l2.rowid < l1.rowid) AS count
      FROM TABLENAME l1) AS t1,
     (SELECT l1.day, l1.price, 
          (SELECT COUNT(*) 
          FROM TABLENAME l2 
          WHERE l2.rowid < l1.rowid) AS count
      FROM TABLENAME l1) AS t2
WHERE t1.count=t2.count-1

This works under the assumption that rowids are always increasing.

Answer 4

Its the same idea as the other, but just uses the fields instead of the rowid. This does exactly what you want:

CREATE TABLE Prices (
    day DATE,
    price FLOAT
);

INSERT INTO Prices(day, price) VALUES(date('now', 'localtime', '+1 day'), 0.5);
INSERT INTO Prices(day, price) VALUES(date('now', 'localtime', '+0 day'), 1);
INSERT INTO Prices(day, price) VALUES(date('now', 'localtime', '-1 day'), 2);
INSERT INTO Prices(day, price) VALUES(date('now', 'localtime', '-2 day'), 7);
INSERT INTO Prices(day, price) VALUES(date('now', 'localtime', '-3 day'), 8);
INSERT INTO Prices(day, price) VALUES(date('now', 'localtime', '-4 day'), 10);

SELECT p1.day, p1.price, p1.price-p2.price 
FROM
    Prices p1, Prices p2,
    (SELECT t2.day AS day1, MAX(t1.day) AS day2 
    FROM Prices t1, Prices t2
    WHERE t1.day < t2.day
    GROUP BY t2.day) AS prev
WHERE p1.day=prev.day1
    AND p2.day=prev.day2

If you want to add the WHERE item='apple' bit you'd add that to both WHERE clauses.