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Rounding Up with given precision Expand / Collapse
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Posted Monday, May 6, 2013 3:20 AM
Grasshopper

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I would like to round up some figures with given precision. So far I can't find a way to do it. ROUND function doesn't work for me because it can't round up, CEILING doesn't work as well because the results is integer.
What I would like to achieve is a rounding up with give precision of 2 places after dot i.e.
5.0000016 should be 5.01
6.1000138 should be 6.11
7.1200073 should be 7.13

Is it any way to do it in SQL?
Post #1449628
Posted Monday, May 6, 2013 3:32 AM
SSCrazy

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Maybe there is a better way, but here is one way:

declare @Num numeric(8,7)
set @Num = 0.0000001

SET @Num = CASE WHEN CAST(@Num AS NUMERIC(3,2)) = @Num THEN @Num ELSE CAST(@Num AS NUMERIC(3,2)) + 0.01 END
SELECT @Num


Adi


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Post #1449630
Posted Monday, May 6, 2013 3:39 AM
Grasshopper

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It works for me. Thank you very much.
Post #1449634
Posted Monday, May 6, 2013 5:25 AM


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Haven't proofed (I normally don't do rounding) it but why wouldn't ROUND(somenumber+.00499999,2,1) work for you?

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Post #1449665
Posted Monday, May 6, 2013 5:39 AM
SSCrazy

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Jeff Moden (5/6/2013)
Haven't proofed (I normally don't do rounding) it but why wouldn't ROUND(somenumber+.00499999,2,1) work for you?


This would work as long as it is guaranteed that that each number will have a digit greater then zero in one of the 2 most left digits after the period (Hope that I wrote it correctly in English). Here is an example:

--with both numbers the user wanted to get 0.01
select ROUND(0.0100000+.00499999,2,1)
select ROUND(0.0000001+.00499999,2,1)


Adi


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Post #1449672
Posted Monday, May 6, 2013 11:10 AM


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Here is another option (The CTE is just how I passed in the sample data):


with TestData as (
select datavalue
from (
values
(5.0000016), -- should be 5.01
(6.1000138), -- should be 6.11
(7.1200073), -- should be 7.13
(5.0000000), -- should be 5.00
(6.1000000), -- should be 6.10
(7.1200000) -- should be 7.12
)dt(datavalue)
)
select
datavalue,
(ceiling(datavalue * 100) * 1.0) / 100
from
TestData;





Lynn Pettis

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Post #1449806
Posted Tuesday, May 7, 2013 9:04 AM


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I came up with another solution based on the same mathematical concept as Lynn's, but I think Lynn's is more elegant:
with TestData as (
select datavalue
from (
values
(5.0000016), -- should be 5.01
(6.1000138), -- should be 6.11
(7.1200073), -- should be 7.13
(5.0000000), -- should be 5.00
(6.1000000), -- should be 6.10
(7.1200000) -- should be 7.12
)dt(datavalue)
)
select
datavalue,
(ceiling(datavalue * 100) * 1.0) / 100, -- Lynn's idea
CASE WHEN (datavalue * 100) % 1 > 0 THEN ROUND(datavalue, 2, 1) + 0.01 ELSE ROUND(datavalue, 2, 1) END -- My idea
from
TestData;



Jason Wolfkill
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Post #1450204
Posted Tuesday, May 7, 2013 10:28 AM
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Here is a solution using the ROUND function:
with test_data_cte as
(
select datavalue
from
(
values
(5.0000016), -- should be 5.01
(6.1000138), -- should be 6.11
(7.1200073), -- should be 7.13
(5.0000000), -- should be 5.00
(6.1000000), -- should be 6.10
(7.1200000) -- should be 7.12
) dt (datavalue)
)
select
datavalue,
rounded_datavalue =
convert(numeric(8,2),round(datavalue+0.0049999,2))
from
test_data_cte

Results:
datavalue                               rounded_datavalue
--------------------------------------- -----------
5.0000016 5.01
6.1000138 6.11
7.1200073 7.13
5.0000000 5.00
6.1000000 6.10
7.1200000 7.12
Post #1450242
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