# Converted Oracle Months_Between to SQL Server Function?

G

#### Gordon

##### Guest
Gordon Asks: Converted Oracle Months_Between to SQL Server Function?
I performed a search here looking for a nice clean solution to the Oracle Months_Between() conversion to SQL Server. I didn't find anything that made sense other than to try to repeat the inaccuracy of the Oracle version. I also searched the web for a little more and found an article on the same thing. Using the code supplied in the article just showed that the guy writing the article didn't know how to test his own code. Let's just say it was scary.

I know this is not so much a question but rather a solution that others might enjoy using. Anyone who has had to use Oracle's Months_Between() function knows that the calculation uses a hard-coded value of 31 and does not include the first day. In some situations, it is nice to have some accuracy. Below is my version of Months_Between() but with 100% accuracy. I still question why Oracle did it that way, or if the Oracle programmer didn't have time to go back and fix the hard coding.

I hope someone out there finds this somewhat useful.

Code:
/*************************************************************************************************
Name    : fsMonthsBetweenDates
Purpose : Mimics the Months_Between function in Oracle; except this one is 100% accurate
Author  : Gordon de Rouyan
Date    : 2022-09-24
*************************************************************************************************/
CREATE OR ALTER FUNCTION dbo.fsMonthsBetweenDates
(
@Date1 datetime,
@Date2 datetime,
@InclusiveFirstDay bit = 1
)
RETURNS float
AS
BEGIN
-- Using all floats for numeric values
DECLARE @Months float = 0;

IF @Date1 IS NOT NULL AND @Date2 IS NOT NULL
BEGIN
-- Include the first day or not; e.g. 2021-02-01 and 2021-02-01
-- Do you want to consider this combination as one day 2021-02-01@00:00:00 to 2021-02-01@23:59:59?
DECLARE @FirstDayInclusive float = 0;

IF @InclusiveFirstDay = 1
SET @FirstDayInclusive = 1;

-- convert early
DECLARE @MbdDate1 date = CAST(@Date1 AS date),
@MbdDate2 date = CAST(@Date2 AS date);

DECLARE @MdbDate1Day float = CAST(DAY(@MbdDate1) AS float),
@MdbDate2Day float = CAST(DAY(@MbdDate2) AS float);

-- Number of days in each month specific to each date; exact with leap year
DECLARE @Date1MonthDays float = CAST(DAY(EOMONTH(@MbdDate1)) AS float),
@Date2MonthDays float = CAST(DAY(EOMONTH(@MbdDate2)) AS float);

-- The fraction of the start month
DECLARE @StartMonthFraction float = CASE
WHEN @MbdDate1 < @MbdDate2 THEN (@Date1MonthDays - @MdbDate1Day + @FirstDayInclusive) / @Date1MonthDays
ELSE (@Date2MonthDays - @MdbDate2Day + @FirstDayInclusive) / @Date2MonthDays
END;

-- The fraction of the end month
DECLARE @EndMonthFraction float =   CASE
WHEN @MbdDate1 < @MbdDate2 THEN @MdbDate2Day / @Date2MonthDays
ELSE @MdbDate1Day / @Date1MonthDays
END;

-- Force positive value returned
SET @Months =   CASE
WHEN @MbdDate1 <= @MbdDate2 THEN DATEDIFF(MONTH, @MbdDate1, @MbdDate2) - 1 + (@StartMonthFraction + @EndMonthFraction)
ELSE DATEDIFF(MONTH, @MbdDate2, @MbdDate1) - 1 + (@StartMonthFraction + @EndMonthFraction)
END;
END;

RETURN @Months;
END;

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SolveForum.com may not be responsible for the answers or solutions given to any question asked by the users. All Answers or responses are user generated answers and we do not have proof of its validity or correctness. Please vote for the answer that helped you in order to help others find out which is the most helpful answer. Questions labeled as solved may be solved or may not be solved depending on the type of question and the date posted for some posts may be scheduled to be deleted periodically. Do not hesitate to share your thoughts here to help others.

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SolveForum.com may not be responsible for the answers or solutions given to any question asked by the users. All Answers or responses are user generated answers and we do not have proof of its validity or correctness. Please vote for the answer that helped you in order to help others find out which is the most helpful answer. Questions labeled as solved may be solved or may not be solved depending on the type of question and the date posted for some posts may be scheduled to be deleted periodically. Do not hesitate to share your thoughts here to help others.

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I would like to start from the last value and create various x values (up to 2500) while the y can stay the same. So that data2 {917, 2} matches the dimensions of data 1 {1337,2}. Any insight would be greatly appreciated.

Code:
 In[1]:= data1 =
Import["/Users/julissavelasquez/Box/1_Harrison Lab/03_Formic \
Acid/Wodtke_2021_Fig4C/hyperthermal_dist.xlsx", "SkipLines" -> 2][[1]];

In[2]:= data2 =
Import["/Users/julissavelasquez/Box/1_Harrison Lab/03_Formic \
Acid/Wodtke_2021_Fig4C/thermal_dist.xlsx", "SkipLines" -> 2][[1]];

In[3]:= {Dimensions[data1], Dimensions[data2]}

Out[3]= {{1337, 2}, {917, 2}}

In[4]:= data2[[1 ;; 2]]

Out[4]= {{2.9895, 0.00054879}, {5.979, 0.00054814}}

In[5]:= data2[[-2 ;; -1]]

Out[5]= {{2487.29, 2.7924*10^-6}, {2490.28, 2.1354*10^-6}}

In[6]:= Table[data2, {i, 2490.28, 2500}]

Out[6]= If[25048064219406400976 === \$SessionID,
Out[6], Message[
MessageName[Syntax, "noinfoker"]]; Missing["NotAvailable"]; Null]

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alwayscurious Asks: How can I Prove this? Is the explanation correct?
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