Maths - Delta Query

Not sure exactly why you want it as sql but it's fairly easy to do....


WITH yourdata
     AS (SELECT 1 id, 25 n FROM DUAL
         UNION ALL
         SELECT 2, 24 FROM DUAL
         UNION ALL
         SELECT 3, 24 FROM DUAL
         UNION ALL
         SELECT 4, 25 FROM DUAL
         UNION ALL
         SELECT 5, 25 FROM DUAL)
SELECT SUM(diff)
  FROM (SELECT n - LAG(n) OVER (ORDER BY id) diff FROM yourdata);



Assuming your data is in a table somewhere, simply remove the WITH clause and change yourdata to your table and the other columns as appropriate

Thanks, But i need Answer for that my formula
Delta means - It is just the difference between any two values.

I know what a delta is

n - LAG(n) OVER (ORDER BY id)   --- there's your delta

but you asked for the sum of them   so sum in the outer query

and that's 0

so that mean my answer is correct?

yes

A variant on that sql that displays the details; the "LAG(N)OVER(ORDERBYID)" column is the n of the previous row.
The column "DIFF" when summed = 0

| LAG(N)OVER(ORDERBYID) |  N |   DIFF |
|-----------------------|----|--------|
|                (null) | 25 | (null) |
|                    25 | 24 |     -1 |
|                    24 | 24 |      0 |
|                    24 | 25 |      1 |
|                    25 | 25 |      0 |

WITH yourdata
      AS (
         SELECT 1 id, 25 n FROM DUAL
         UNION ALL
         SELECT 2, 24 FROM DUAL
         UNION ALL
         SELECT 3, 24 FROM DUAL
         UNION ALL
         SELECT 4, 25 FROM DUAL
         UNION ALL
         SELECT 5, 25 FROM DUAL
            )
SELECT
      LAG(n) OVER (ORDER BY id)
    , n
    , n - LAG(n) OVER (ORDER BY id) diff
FROM yourdata
;

http://sqlfiddle.com/#!4/d41d8/33755

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