Find nearest SUM amount from table SQL SERVER - sql

Question

Query
Declare #table1 TABLE (accountno varchar(max), saved_amount decimal)
INSERT INTO #table1 VALUES
('001',25),
('002',5)
Declare #table2 TABLE (accountno varchar(max), payamount decimal,ilno int)
INSERT INTO #table2 VALUES
('001',10,1),
('001',10,2),
('001',10,3),
('001',10,4),
('002',10,1),
('002',10,2);
WITH aa
AS (
SELECT a.*
,b.ilno
,b.payamount
,SUM(payamount) OVER (
PARTITION BY a.accountno ORDER BY CAST(a.accountno AS INT)
,ilno
) AS total_amount
FROM #table1 a
LEFT JOIN #table2 b ON a.accountno = b.accountno
)
,bb
AS (
SELECT accountno
,MAX(ilno) AS ilno
FROM aa
WHERE saved_amount >= total_amount
GROUP BY accountno
)
SELECT a.* FROM aa a INNER JOIN bb b on a.accountno =b.accountno AND a.ilno = b.ilno
Result
accountno | saved_amount | ilno | payamount | total_amount
----------------------------------------------------------
001 | 25 | 2 | 10 | 20
Expected Result
accountno | saved_amount | ilno | payamount | total_amount
----------------------------------------------------------
001 | 25 | 2 | 10 | 20
002 | 5 | 1 | 10 | 10
What I want is
If saved_amount is less than the first ilno, then get the first ilno else
get the highest ilno where saved_amount>=total_amount

Answer 1

You have a running total that you compare with the saved amount. You want the highest running total that doesn't exceed the saved amount. But in case even the initial pay amount exceeds the saved amount already, you want to default to this record. So the main task is to find a way of ranking the records. In my query I do it like this:
Prefer records where the running total does not exceed the saved amount.
Then look at the abolute of their difference and take the smallest.
There are certainly other ways that achieve the same. Maybe even methods that you find more readable. Then just adjust the order by clause in the ranking query.
with summed as
(
select
t1.*,
from #table1 t1
join
(
select
ilno,
payamount,
sum(payamount) over (partition by accountno order by ilno) as total_amount
from #table2
) on t2.accountno = t1.accountno
)
, ranked as
(
select summed.*,
row_number() over (partition by accountno
order by case when saved_amount >= total_amount then 1 else 2 end,
abs(saved_amount - total_amount)
) as rn
)
select *
from ranked
where rn = 1;
This is not the "nearest sum", as you said in the title, but the one that obeys the specified rules. So with a saved amount of 100 and paid amounts of first 1 and then 100, you'd get the record with a total of 1 (which is 99 less than the saved amount) and not the one with a total of 101 (which is only 1 more than the saved amount).

Answer 2

Other way to solve using flags:
first calculated one flag to point if saved_amount >= payamount for current row
calculated three more flags:
group_flag to show is there a case where saved_amount >= payamount for the given accountno
[min_ilno] and [max_ilno] for given account
Having this flags, the final result set is calculated easily. Here is the code:
WITH DataSource AS
(
SELECT a.*
,b.ilno
,b.payamount
,SUM(payamount) OVER (PARTITION BY a.accountno ORDER BY ilno) AS total_amount
,IIF(a.saved_amount >= SUM(payamount) OVER (PARTITION BY a.accountno ORDER BY ilno), 1, 0) AS [flag]
FROM #table1 a
LEFT JOIN #table2 b
ON a.accountno = b.accountno
),
DataSourceFinal AS
(
SELECT *
,MAX(flag) OVER (PARTITION BY accountno) as [group_flag]
,MIN(IIF(flag = 0 ,ilno, NULL)) OVER (PARTITION BY accountno) as [min_ilno]
,MAX(IIF(flag = 1 ,ilno, NULL)) OVER (PARTITION BY accountno) as [max_ilno]
FROM DataSource
)
SELECT accountno, saved_amount, ilno, payamount, total_amount
FROM DataSourceFinal
WHERE ([group_flag] = 1 AND [ilno] = [max_ilno])
OR ([group_flag] = 0 AND [ilno] = [min_ilno]);
and the output: