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Rounding Up with given precision


Rounding Up with given precision

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phoenix_
phoenix_
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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?
Adi Cohn
Adi Cohn
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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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phoenix_
phoenix_
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It works for me. Thank you very much.
Jeff Moden
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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?

--Jeff Moden

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Adi Cohn
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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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Lynn Pettis
Lynn Pettis
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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;




Cool
Lynn Pettis

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wolfkillj
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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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Michael Valentine Jones
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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

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