Question

I’ve got a mysql table that has a running total:

+---------------------+--------+
| Timestamp           | Total  |
+---------------------+--------+
| 2012-07-04 05:35:00 |  1.280 | 1.280-1.280 = 0
| 2012-07-04 09:25:00 |  2.173 | 2.173-1.280 = 0.893
| 2012-07-04 09:30:00 |  2.219 | 2.219-1.280 = 0.939
| 2012-07-04 15:00:00 |  7.778 | 7.778-1.280 = 6.498
| 2012-07-04 21:05:00 | 13.032 | 13.032-1.280 = 11.752
| 2012-07-04 22:00:00 | 13.033 | 13.033-1.280 = 11.753
| 2012-07-05 05:20:00 | 13.033 | 13.033-13.033 = 0
| 2012-07-05 07:10:00 | 13.140 | 13.140-13.033 = 0.107
| 2012-07-05 10:15:00 | 14.993 | 14.993-13.033 = 1.960
| 2012-07-05 11:35:00 | 16.870 | 16.870-13.033 = 3.837
+---------------------+--------+

What I’m looking for is a query that determines the aggregated daily increase for each interval.
I’ve tried to show the desired outcome as well as the calculation behind each row. I’ve tried already several things with a join, but somehow I fail to determine what the starting value for each day is.

Thanks.

Answer 1

I can’t vouch for the efficiency of this query, but it does get you the results you are looking for:

SELECT t1.`Timestamp`, t1.`Total`, 
      CASE WHEN t1.`timestamp` =
            (SELECT MIN(t2.`Timestamp`)
             FROM myTable t2 
             WHERE DATE(t2.`Timestamp`)=DATE(t1.`Timestamp`)) 
        THEN 0
        ELSE t1.`Total` - (SELECT MIN(t3.`Total`)
              FROM myTable t3
              WHERE DATE(t3.`Timestamp`)=DATE(t1.`Timestamp`))         
        END AS Diff
FROM myTable t1
ORDER BY `Timestamp`

Alternate Solution (more efficient I think)

SELECT  t1.`Timestamp`, t1.`Total`, (t1.`Total` - d1.MinVal) diff
FROM myTable t1 
  INNER JOIN 
     (SELECT DATE(`Timestamp`) ts_date,
             MIN(`Total`) AS MinVal
      FROM myTable
      GROUP BY ts_date) d1
  ON DATE(t1.`Timestamp`) = d1.ts_date