How to SUM() each row into another column - sql

Question

I have this table
| ID_prim | ID (FKey) | Date | Moved Items |
|:-----------|:------------|-------------:|:------------:|
| 1003 | 12_1 | nov 2013 | 2 |
| 1003 | 12_2 | okt 2013 | 3 |
| 1003 | 12_3 | dec 2014 | 5 |
| 1003 | 12_4 | feb 2015 | 10 |
| 1003 | 12_5 | apr 2012 | 1 |
| 1003 | 12_11 | jan 2011 | 5 |
I want to query the same table as follows:
Order the Date by desc
Sum each 'Moved Item" per row
Stop the query if the Sum reaches my desired amount
My desired amount starts from the MAX 'Summed Total' (26) and subtracts the amount I want (16)
Like so
| ID_prim | ID (FKey) | Date | Moved Items | Summed Total |
|:-----------|:------------|-------------:|:------------:|:------------:|
| 1003 | 12_4 | feb 2015 | 10 | 26
| 1003 | 12_3 | dec 2014 | 5 | 16
| 1003 | 12_3 | nov 2013 | 2 | 11 <
| 1003 | 12_4 | okt 2013 | 3 | 9
| 1003 | 12_5 | apr 2012 | 1 | 6
| 1003 | 12_11 | jan 2011 | 5 | 5
I want to stop the query when i reach "Summed Total" (26) - 16 = 10. So Show me everything from 10 >
I would only get these values in the database.
| ID_prim | ID (FKey) | Date | Moved Items | Summed Total |
|:-----------|:------------|-------------:|:------------:|:------------:|
| 1003 | 12_4 | feb 2015 | 10 | 26
| 1003 | 12_3 | dec 2014 | 5 | 16
| 1003 | 12_3 | nov 2013 | 2 | 11
What I have is the following
SELECT
T1.ID_prim, T1.ID as ID (FKey), T1.Moved_Items as Moved Items, t1.Date, SUM(T2.MOVEMENTQTY) AS Summed Total
FROM Table1 T1
INNER JOIN Table1 T2 ON T2.ID <= T1.ID
inner join table2 inout on T1.ID_prim = inout.ID_prim
AND T2.ID_prim = inout.ID_prim
AND T2.ID_prim = T1.ID_prim
where t1.ID_prim = 1003
and t2.ID_prim = 1003
and inout.ISSOTRX = 'N'
GROUP BY T1.ID_prim, T1.Moved Items, t1.Date
HAVING SUM(T2.Moved Items) <= 16
order by t1.UPDATED desc
But the sum doesn't really work.
Can anyone help me out to make the SQL statement for Oracle DB that will print my Desired table?

Answer 1

Based on OP's clarifications via comments on the question, it could be done using SUM() analytic function to get the running total, and then filter it based on the condition.
Table:
SQL> SELECT * FROM t;
ID_PRIM ID DT MOVED
---------- ----- --------- ----------
1003 12_1 01-NOV-13 2
1003 12_2 01-OCT-13 3
1003 12_3 01-DEC-14 5
1003 12_4 01-FEB-15 10
1003 12_5 01-APR-12 1
1003 12_11 01-JAN-11 5
6 rows selected.
SQL>
Running total
SQL> SELECT t.*, SUM(moved) OVER(ORDER BY dt) sm FROM t ORDER BY dt DESC;
ID_PRIM ID DT MOVED SM
---------- ----- --------- ---------- ----------
1003 12_4 01-FEB-15 10 26
1003 12_3 01-DEC-14 5 16
1003 12_1 01-NOV-13 2 11
1003 12_2 01-OCT-13 3 9
1003 12_5 01-APR-12 1 6
1003 12_11 01-JAN-11 5 5
6 rows selected.
SQL>
Desired output
SQL> WITH DATA AS
2 ( SELECT t.*, SUM(moved) OVER(ORDER BY dt) sm FROM t ORDER BY dt DESC
3 )
4 SELECT * FROM data WHERE sm >= 16;
ID_PRIM ID DT MOVED SM
---------- ----- --------- ---------- ----------
1003 12_4 01-FEB-15 10 26
1003 12_3 01-DEC-14 5 16
SQL>
Please note that, nov 2013 is not a date, it is a string. Since you want to sort on the basis of date, you must always use TO_DATE to explicitly convert it into date. Anyway, I used TO_DATE to create the sample data.
Update OP wants to subtract his desired value from the MAX value of the summed up values at run time.
SQL> WITH DATA AS
2 ( SELECT t.*, SUM(moved) OVER(ORDER BY dt) sm FROM t ORDER BY dt DESC
3 )
4 SELECT * FROM DATA t WHERE sm >
5 (SELECT MAX(sm) FROM data
6 ) - 16 ;
ID_PRIM ID DT MOVED SM
---------- ----- --------- ---------- ----------
1003 12_4 01-FEB-15 10 26
1003 12_3 01-DEC-14 5 16
1003 12_1 01-NOV-13 2 11
SQL>
In the updated query, MAX(sm) returns 26, and then the rows are filtered on the condition WHERE sm > MAX(sm) -16 which means return all the rows where the 'sm' value is greater than 26 -16 i.e. 10. You could use a substitution variable to input the value 16 at run time.