Custom rounding in SQL - sql-server

Question

How can I set rules for custom rounding in SQL?
I need set this rules:
if second number after decimal is 1-2 eg. 10,02 - round it to 0 result 10,00
if second number after decimal is 3-7 eg. 10,13 - round it to 5 result 10,15
if second number after decimal is 8-9 eg. 10,28 - round it to bigger number result 10,3
Can someone help me how to do this? Round function works from 0-5 and above 5 but how to do "custom rounding function" based on this rules?
Thanks in advance.

Answer 1

One possible approach is the following statement:
SELECT
[Number],
CASE
WHEN ([Number] * 100) % 10 BETWEEN 3 AND 7 THEN ROUND([Number] + 0.02, 1) - 0.05
ELSE ROUND([Number] + 0.02, 1)
END [RoundedNumber]
FROM (VALUES (10.00), (10.01), (10.12), (10.13), (10.14), (10.28), (10.29)) v ([Number])
Result:
Number RoundedNumber
10.00 10.00
10.01 10.00
10.12 10.10
10.13 10.15
10.14 10.15
10.28 10.30
10.29 10.30

Answer 2

You can create your own function to use your custom rounding rules:
CREATE FUNCTION CustomRound(#num DECIMAL(18,2))
RETURNS DECIMAL(18,2)
AS
BEGIN
RETURN ROUND(#num, 1, 1) +
CASE WHEN (#num - ROUND(#num, 1, 1)) * 100 BETWEEN 1 AND 2 THEN 0
WHEN (#num - ROUND(#num, 1, 1)) * 100 BETWEEN 3 AND 7 THEN 0.05
WHEN (#num - ROUND(#num, 1, 1)) * 100 BETWEEN 8 AND 9 THEN 0.1
END
END
This function truncates the decimal number after first decimal place and add depending on the second decimal place a value to "round" as needed.
You can use the above function like this:
SELECT dbo.CustomRound(10.12) -- 10.10
SELECT dbo.CustomRound(10.02) -- 10.00
SELECT dbo.CustomRound(10.13) -- 10.15
SELECT dbo.CustomRound(10.28) -- 10.30
demo on dbfiddle.uk
You can also extend this function to be more dynamic using the custom rounding:
CREATE FUNCTION CustomRound(#num DECIMAL(18,6), #precision INT)
RETURNS DECIMAL(18,6)
AS
BEGIN
DECLARE #prec INT = IIF(#precision > 0, #precision - 1, 0);
RETURN ROUND(#num, #prec, 1) +
CASE WHEN (#num - ROUND(#num, #prec, 1)) * POWER(10, #precision) BETWEEN 1 AND 2 THEN 0.0 / POWER(10, #precision)
WHEN (#num - ROUND(#num, #prec, 1)) * POWER(10, #precision) BETWEEN 3 AND 7 THEN 5.0 / POWER(10, #precision)
WHEN (#num - ROUND(#num, #prec, 1)) * POWER(10, #precision) BETWEEN 8 AND 9 THEN 10.0 / POWER(10, #precision)
END
END
You can use this function like this:
SELECT dbo.CustomRound(10.12, 2) -- 10.10
SELECT dbo.CustomRound(10.02, 2) -- 10.00
SELECT dbo.CustomRound(10.13, 2) -- 10.15
SELECT dbo.CustomRound(10.28, 2) -- 10.30
-- or
SELECT dbo.CustomRound(10.102, 3) -- 10.100
SELECT dbo.CustomRound(10.002, 3) -- 10.000
SELECT dbo.CustomRound(10.103, 3) -- 10.105
SELECT dbo.CustomRound(10.208, 3) -- 10.210
demo on dbfiddle.uk

Answer 3

You can define your custome funtion and use it for ever
DECLARE #NUM REAL =10.01;
DECLARE #RESULT NVARCHAR(MAX)
IF #NUM=ROUND(#NUM,0)
BEGIN
SET #RESULT=#NUM
END
ELSE
BEGIN
DECLARE #NUM_STR VARCHAR(MAX)=CAST(#NUM AS VARCHAR(MAX)) -- CONVERT TO VARCHAR
DECLARE #LEFT VARCHAR(MAX)= CASE WHEN #NUM>=0 THEN FLOOR(#NUM) ELSE CEILING(#NUM) END -- GET LEFT SIDE NUMBERS
DECLARE #RIGHT VARCHAR(MAX)= REPLACE(#NUM_STR,CONCAT(#LEFT,'.'),'') -- GET RIGHT SIDE NUMBERS
... -- YOUR ROLES
END

Answer 4

You can multiply by 20, then round to the whole number and divide by 20 again.
e.g.
SELECT ROUND(myNum * 20,0) / 20.0
FROM myTable;