How can I calculate the remaining amount per row? - sql

Question

I have a table that I want to find for each row id the amount remaining from the total. However, the order of amounts is in an ascending order.
id amount
1 3
2 2
3 1
4 5
The results should look like this:
id remainder
1 10
2 8
3 5
4 0
Any thoughts on how to accomplish this? I'm guessing that the over clause is the way to go, but I can't quite piece it together.Thanks.

Answer 1

Since you didn't specify your RDBMS, I will just assume it's Postgresql ;-)
select *, sum(amount) over() - sum(amount) over(order by amount) as remainder
from tbl;
Output:
| ID | AMOUNT | REMAINDER |
---------------------------
| 3 | 1 | 10 |
| 2 | 2 | 8 |
| 1 | 3 | 5 |
| 4 | 5 | 0 |
How it works: http://www.sqlfiddle.com/#!1/c446a/5
It works in SQL Server 2012 too: http://www.sqlfiddle.com/#!6/c446a/1
Thinking of solution for SQL Server 2008...
Btw, is your ID just a mere row number? If it is, just do this:
select
row_number() over(order by amount) as rn
, sum(amount) over() - sum(amount) over(order by amount) as remainder
from tbl
order by rn;
Output:
| RN | REMAINDER |
------------------
| 1 | 10 |
| 2 | 8 |
| 3 | 5 |
| 4 | 0 |
But if you really need the ID intact and move the smallest amount on top, do this:
with a as
(
select *, sum(amount) over() - sum(amount) over(order by amount) as remainder,
row_number() over(order by id) as id_sort,
row_number() over(order by amount) as amount_sort
from tbl
)
select a.id, sort.remainder
from a
join a sort on sort.amount_sort = a.id_sort
order by a.id_sort;
Output:
| ID | REMAINDER |
------------------
| 1 | 10 |
| 2 | 8 |
| 3 | 5 |
| 4 | 0 |
See query progression here: http://www.sqlfiddle.com/#!6/c446a/11

Answer 2

I just want to offer a simpler way to do this in descending order:
select id, sum(amount) over (order by id desc) as Remainder
from t
This will work in Oracle, SQL Server 2012, and Postgres.
The general solution requres a self join:
select t.id, coalesce(sum(tafter.amount), 0) as Remainder
from t left outer join
t tafter
on t.id < tafter.id
group by t.id

Answer 3

SQL Server 2008 answer, I can't provide an SQL Fiddle, it seems it strips the begin keyword, resulting to syntax errors. I tested this on my machine though:
create function RunningTotalGuarded()
returns #ReturnTable table(
Id int,
Amount int not null,
RunningTotal int not null,
RN int identity(1,1) not null primary key clustered
)
as
begin
insert into #ReturnTable(id, amount, RunningTotal)
select id, amount, 0 from tbl order by amount;
declare #RunningTotal numeric(16,4) = 0;
declare #rn_check int = 0;
update #ReturnTable
set
#rn_check = #rn_check + 1
,#RunningTotal =
case when rn = #rn_check then
#RunningTotal + Amount
else
1 / 0
end
,RunningTotal = #RunningTotal;
return;
end;
To achieve your desired output:
with a as
(
select *, sum(amount) over() - RunningTotal as remainder
, row_number() over(order by id) as id_order
from RunningTotalGuarded()
)
select a.id, amount_order.remainder
from a
inner join a amount_order on amount_order.rn = a.id_order;
Rationale for guarded running total: http://www.ienablemuch.com/2012/05/recursive-cte-is-evil-and-cursor-is.html
Choose the lesser evil ;-)