> Something like this:
>
> select id,db,cr,(select sum(cr-db) from calc sub where sub.id <=
> calc.id) from calc;
>
> This of course assumes that ID indicates the correct order of the
> entries and it will blow up if you allow nulls for cr or db (which
> you shouldn't since that would literally be interpreted as "they
> withdrew 10 and deposited an unknown amount"). If you have null
> values already and they should be interpreted as 0 just do this:
>
> select id, db, cr, (select sum(coalesce(cr,0)-coalesce(db,0)) from
> calc sub where sub.id <= calc.id) from calc;
>
> I assume no responsibility for potential lack of scalability of this
> query. :) It's quite possible a faster solution exists - we'll see
> what comes up on the list.
>
If data from "View" without ID, how can I do?
My View:
trx_date | trx_time | descriptions |
payment_method | debet | credit | creator
------------+----------+--------------------------------------+-------------
---+-------+----------+---------
2003-10-09 | 21:55:02 | Resto Biling : 13,800, Paid : 10,000 | Visa
| 3800 | | middink
2003-10-16 | 03:28:30 | Payment - Thank You | Visa
| | 40000.00 | middink
2003-10-08 | 18:17:40 | Payment - Thank You | Cash
| | 50000.00 | middink
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